Pearson Physics Level 20 Unit I Kinematics: Chapter 1 Solutions

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1 Pearson Physics Level 0 Unit I Kinematics: Chapter 1 Solutions Student Book page 9 Skills Practice 1. scale: 6.0 m : 3.05 cm (north/south side of rink) scale: 60.0 m : 7.05 cm (east/west side of rink) (a) position from north side of rink: position from south side of rink: player 1: 0.55 cm = 4.7 m [S] player 1:.50 cm = 1.3 m [N] player : 0.75 cm = 6.4 m [S] player :.30 cm = 19.6 m [N] player 3:.45 cm = 0.9 m [S] player 3: 0.55 cm = 4.7 m [N] player 4:.60 cm =. m [S] player 4: 0.40 cm = 3.4 m [N] player 5:.0 cm = 18.8 m [S] player 5: 0.90 cm = 7.7 m [N] (b) position from east side of rink: player 1: 5.00 cm = 4.6 m [W] player : 3.70 cm = 31.5 m [W] player 3:.15 cm = 18.3 m [W] player 4: 3.80 cm = 3.3 m [W] player 5: 6.85 cm = 58.3 m [W] (c).0 cm = 17.0 m [S] position from west side of rink: player 1:.00 cm = 17.0 m [E] player : 3.30 cm = 8.1 m [E] player 3: 4.90 cm = 41.7 m [E] player 4: 3.0 cm = 7. m [E] player 5: 0.0 cm = 1.7 m [E] Example 1.1 Practice Problems 1. Given d m [N] d 0.0 m [N] d m [N] displacement ( d ) Since the sprinter moves north continuously, the distances can be added together. d = 40.0 m [N] m [N] m [N] = m [N] Sprinter s displacement is m [N].. Given d m [right] d 3.50 m [left] displacement ( d ) 1 Copyright 009 Pearson Education Canada

2 Use vector addition, but change signs since directions are opposite. d = 0.75 m [right] m [left] = m [left] m [left] =.75 m [left] Player s displacement is.75 m [left]. 3. Given d m [back] d 0.85 m [forth] distance ( d ) displacement ( d ) The bricklayer s hand moves 1.70 m back and forth four times, so d 4( d1 d). d 4(0.85 m 0.85 m) 6.80 m Since the player starts and finishes in the same spot, displacement is zero. d 0 m Total distance is 6.80 m. Total displacement is zero. Student Book page Check and Reflect Knowledge 1. Two categories of terms that describe motion are scalar quantities and vector quantities. Scalar quantities include distance, time, and speed. Vector quantities include position, displacement, velocity, and acceleration.. Distance is the length of path taken to travel between two points. Displacement is the change in position: how far and in which direction an object is situated after it has travelled from its original starting or reference point. Distance is a scalar quantity, represented by d. Displacement is a vector quantity, represented by d. Distance may be the same as displacement for straight line motion in one direction, but will be different from displacement when the motion is along any path for which the direction of motion changes. 3. A reference point determines the direction for vector quantities. A reference point is necessary to calculate displacement and measure position. Copyright 009 Pearson Education Canada

3 Applications 4. (a) (d) (e) From Greg to Hannah, the displacement is 7.5 m [right] m [right] m [right] = 13.0 m [right]. 5. Given d 50.0 km [W] d i 5.0 km [E] final position ( d f ) Use the equation d df di. d df di d dd f i d f 50.0 km [W] 5.0 km [E] 50.0 km [W] 5.0 km [W] 45.0 km [W] The person s final position is 45.0 km [W]. 6. Given d1 3.0 m [left] d (3.0 m 5.0 m) [right] distance ( d ) displacement ( d ) Use D d = d1+ d D d = d1+ d d = 3.0 m + ( ) m = 11.0 m d 3.0 m [left] (3.0 m 5.0 m)[right] 3.0 m [right] 8.0 m[right] 5.0 m [right] 3 Copyright 009 Pearson Education Canada

4 The ball travels a distance of 11.0 m. Its displacement is 5.0 m [right]. 7. d groom = 0.50 m [right] d best man = 0.75 m [left] d maid of honour = 0.50 m [right] m [right] = 1.5 m [right] d flower girl = 0.75 m [left] m [left] = 1.50 m [left] Student Book page 13 Concept Check (a) On a ticker tape at rest, all the dots would be placed on top of each other at a single point. The slope of the position-time graph in Figure 1.14 is zero. (b) A position-time graph for an object travelling at a constant velocity is a straight line. Its slope is positive (tilted to the left) for positive velocity, negative (tilted to the right) for negative velocity, and zero (horizontal) for an object at rest. In each case, change in position remains constant for equal time intervals. Student Book page 14 Concept Check The velocity of the ball can be positive with the hole at the origin if the slope of the graph is positive. The axis is labelled right and the initial position of the ball is 5.0 m. This means the ball is 5.0 m LEFT of the origin. The graph slopes up from -5.0 m to zero and hence has a positive slope. Example 1. Practice Problems 1.(a) Student Book page 15 4 Copyright 009 Pearson Education Canada

5 d (b) Elk: v = t 00 m [forward] = 10.0 s = 0.0 m/s [forward] d Coyote: v = t 00 m [forward] = 10.4 s = 19. m/s [forward] d Grizzly Bear: v = t 00 m [forward] = 18.0 s = 11.1 m/s [forward] d Moose: v = t 00 m [forward] = 1.9 s = 15.5 m/s [forward] Student Book page 16 Concept Check The position-time graph showing the motion of two objects approaching each other would consist of two converging lines. Example 1.3 Practice Problems 1. (a) 5 Copyright 009 Pearson Education Canada

6 (b) From the graph, B catches up with A at t = 4.0 s. B s position at this time is 0.0 m [right]. Because B started 10.0 m to the left of A, B s displacement is 0.0 m [right] m [right] = 30.0 m [right]. Student Book page 18 Example 1.4 Practice Problem 1. Given Consider right to be positive. da m [right] = m ta= 15.0 s db m [right] = m tb= 10.0 s velocities of A and B ( A, B) d The velocities of each rollerblader are given by. t d A v A t A m 15.0 s 6.67 m/s 6.67 m/s [right] v B d B t B m 10.0 s 10.0 m/s 10.0 m/s [right] 6 Copyright 009 Pearson Education Canada

7 The velocity of rollerblader A is 6.67 m/s [right] and the velocity of rollerblader B is 10.0 m/s [right]. Student Book page 0 1. Check and Reflect Knowledge 1. The quantities of motion that remain the same over equal time intervals for an object at rest are position (the object does not move), displacement (since the object does not move, the change in position is always zero), velocity (velocity is zero), and acceleration (acceleration is zero) although acceleration is not dealt with until after p. 3.. For an object travelling at a constant velocity, displacement is the same over equal time intervals. 3. The faster the ticker tape, the fewer dots there are, and the steeper the graph is. Therefore: (i) D (ii) C (iii) A (iv) B 4. Given A 3.0 m/s B 1.5 m/s db i 0.0 m t = 0.0 s distance between A and B and who is ahead at 0.0 s da db A B t t da At db Bt (3.0 m/s)(0.0 s) (1.5 m/s)(0.0 s) 60 m 30 m But, in total, B is 30 m m away from the origin. Therefore, db 50 m. The distance between A and B is 60 m 50 m = 10 m, and A is ahead. A is ahead of B by 10 m. 7 Copyright 009 Pearson Education Canada

8 5. Given d1 16 km [E] d 3 km [W] final position ( d f ) df = d1+ d = 16 km [E] + 3 km [W] = 16 km [E] + (-3 km [E]) = 16 km [E]-3 km [E] =-7 km [E] = 7 km [W] The camper s final position with respect to the camp site is 7 km [W] (a) (b) 8 Copyright 009 Pearson Education Canada

9 (c) Applications 8. Given va 5.0 m/s [right] vb 4.5 m/s [right] displacement ( d AB ) Determine the displacement of both children. Equation for child A: ya 5.0 t Equation for child B: yb 4.5 t Find the difference between results. Since Child A pedals faster, Child A should be farther away from the initial starting position. ya yb 5.0t 4.5t m 0.5 [right] 5.0 s s.5 m [right] and Verify Child A will be.5 m farther right after 5.0 s. Check: Child A travels 5 m and Child B travels.5 m. 9. Given va 5.0 m/min vb 9.0 cm/s = 5.4 m/min (converted from 9.0 cm/s 0.09 m/s 60 s/min) t 3.0 min Which insect is ahead and by how much after 3.0 min After 3.0 min, A has travelled 5.0 m/min 3.0 min = 15.0 m, and B has travelled 5.4 m/min 3.0 min = 16. m. So the distance between A and B is 1. m. Insect B is ahead. Insect B is ahead by 1. m. 10. A: The object is mong west with a constant speed. B: The object is stationary. C: The object is mong east with a constant speed, slower than in A. 9 Copyright 009 Pearson Education Canada

10 11. Given Choose east to be positive and the mosquito s initial position be the reference point ( dm 0 m [E] ). i vm.4 km/h [E] vy.0 m/s [W] d y 35.0 m [E] (you are 35.0 m east of the mosquito at start) 1 time ( t ) at point of collision distance you travelled before mosquito hit you First, convert.4 km/h to m/s. km 1000 m 1 h m.4 h 1 km 3600 s m/s Position of mosquito at point of collision, dm d f m i mt t m [E] Your position at point of collision, dy d f y i yt 35.0 m [E].0 t m [W] Since d m f and d y f are the same position, t m [E] 35.0 m [E].0 t m [W] 0.667t t.667t t s s Distance you travelled before the mosquito hit you, yt.0 m/s13.1 s 6 m and Verify The mosquito will hit you at 13 s, when you have travelled 6 m toward the mosquito. Check: Mosquito travels m/s 13.1 s = 8.75 m in 13.1 s; 8.75 m + 6. m = 35.0 m, the original separation. 1. Given you: v=.5 m/s [N] di = 0 m friend: v=.0 m/s [N] di = 5.0 m [N] time ( D t ) displacement ( d ) 10 Copyright 009 Pearson Education Canada

11 Equation for you: ya (.5 m/s [N]) t Equation for friend: yb (.0 m/s [N]) t 5.0 m [N] At intersection, ya = yb = d f (.5 m/s [N]) t (.0 m/s [N]) t5.0 m [N] (0.5 m/s [N]) t 5.0 m [N] t 0 s ya (.5 m/s [N])(0 s) 45 m [N] and Verify It takes 0 s to close the gap. Your displacement is 45 m [N]. Check: yb (.0 m/s [N])(0 s) 5.0 m [N] 45 m [N] 13. Given v = 35 km/h [W] p = 300 m [E] (reference point: traffic light) v = 40 km/h [E] A A times for: vehicles to pass each other vehicle A to pass traffic light vehicle B to pass traffic light km 1000 m 1 h va 35 [W] h 1 km 3600 s 9.7 m/s [W] da 300 m [E] 300 m [W] Equation for vehicle A: ya 9.7t 300 km 1000 m 1 h vb 40 [E] h 1 km 3600 s 11.1 m/s [W] db 450 m [W] 300 m [W] 150 m [W] Equation for vehicle B: yb 11.1t 150 Vehicles pass when ya yb. Vehicle A passes traffic light when ya 0. Vehicle B passes traffic light when yb t t t 450 t s 9.7t t 300 t 31 s B 11 Copyright 009 Pearson Education Canada

12 11.1t t 150 t 14 s The vehicles pass after s. Vehicle A passes the traffic light after 31 s. Vehicle B passes the traffic light after 14 s. Student Book page Concept Check (a) The slope of a position-time graph represents velocity. (b) The slope of a velocity-time graph represents acceleration. Student Book page 3 Concept Check The lower ticker tape in Figure 1.5 represents accelerated motion because the spaces between the dots are changing the dots are successively farther apart showing increasing displacement in equal time intervals. The speed is increasing. Student Book page 6 Concept Check The position-time graph for an object undergoing negative acceleration in the positive direction is a parabola that curves down to the right. The ticker tape of the motion of an object that is slowing down would consist of a series of dots that get closer and closer together. Example 1.5 Practice Problems 1. Student Book page 7. Acceleration is the slope of the graph. 1 Copyright 009 Pearson Education Canada

13 a v - v f i = tf - ti = 0 m/s -10 m/s 10 s - 0 s =-1.0 m/s [N] Student Book page 9 Concept Check (a) For a cyclist coming to a stop at a red light, her velocity is positive and her acceleration is negative (in opposite directions). For the space shuttle taking off, its velocity and acceleration are both positive (in the same direction). (b) An object can have a negative acceleration and be speeding up if its velocity is increasing in the negative direction. When velocity and acceleration are in the same direction, an object speeds up. (c) If the positive slopes of the tangents along the curve increase, then the object is speeding up. Similarly, if the negative slopes of the tangents along the curve become more negative (decrease), then the object is speeding up in the negative direction. If the positive slopes decrease or the negative slopes become less negative (increase), then the object is slowing down. Student Book page Check and Reflect Applications.80 m/s [forward] 0.00 m/s [forward] 1. (a) a 0.50 s 0.00 s = 5.6 m/s [forward] 9.80 m/s [forward].80 m/s [forward] (b) a 3.00 s 0.50 s =.8 m/s [forward] m/s [forward] m/s [forward] (c) a 6.00 s 5.00 s = 0.30 m/s [forward] (d) Velocity is increasing whereas acceleration is decreasing.. The object is accelerating (speeding up) to the left. 3. (i) A (ii) B (iii) C (iv) D Extensions 4. d 1 m [forward] 0.0 m [forward] At time.0 s, slope = 3.8 m/s [forward] t 4.0 s 0.8 s d 4 m [forward] 3 m [forward] At time 4.0 s, slope = 7.0 m/s [forward] t 5.0 s.0 s At time 6.0 s, slope = 0.0 m/s [forward] d 9 m [forward] 33 m [forward] At time 8.0 s, slope = 7.5 m/s [forward] t 9. s 6.0 s 13 Copyright 009 Pearson Education Canada

14 Time (s) Velocity (m/s [forward]) Student Book page 3 Concept Check 1. (a) The object is stopped at a positive position. The object is stopped at a negative position. (b) The object is travelling at a constant velocity in the positive direction. The object is travelling at a constant velocity in the negative direction. (c) The object is speeding up while mong in the positive direction. The object is speeding up while mong in the negative direction. The object is slowing down while mong in the positive direction. The object is slowing down while mong in the negative direction.. (a) The slope of the position-time graph is zero, so the velocity-time graph is along the time axis. The slope of the position-time graph is zero, so the velocity-time graph is along the time axis. 14 Copyright 009 Pearson Education Canada

15 (b) The slope of the position-time graph is positive, so the velocity-time graph is a horizontal line above the time axis indicating a constant velocity in the positive direction. The slope of the position-time graph is negative, so the velocity-time graph is a horizontal line below the time axis indicating a constant velocity in the negative direction. (c) The slope of the position-time graph increases in the positive direction, so the velocity-time graph is a straight line with a positive slope above the time axis. Concept Check The slope of the position-time graph increases in the negative direction, so the velocity-time graph is a straight line with a negative slope below the time axis. The slope of the position-time graph decreases in the positive direction, so the velocity-time graph is a straight line with a negative slope above the time axis. The slope of the position-time graph decreases in the negative direction, so the velocity-time graph is a straight line with a positive slope below the time axis. The acceleration-time graphs encountered thus far are all (a) either a zero line (along the time axis), meaning no change in speed or constant motion, or (b) a horizontal line either above or (c) below the time axis. These last two cases represent situations where the object is either speeding up or slowing down, depending on the direction of velocity. (a) (b) (c) Student Book page 34 Example 1.6 Practice Problems 1. D d = area under graph m =. [N] 10 s s = m [N] 15 Copyright 009 Pearson Education Canada

16 . a = slope of curve = 0 m/s [N] 1 æ m ö D d = area under graph = -0 [E] 10 s ç s =-100 m [E] or 100 m [W] çè ø ( ) = m [E] or m [W] Student Book page 36 Concept Check The ball s net displacement is zero because the sum of the areas under the velocity-time graph is zero. That is, the ball returns to the position from which it was thrown. Student Book page 37 Example 1.7 Practice Problems 1. Given Consider east to be positive. d 1 = 10.0 m [E] = 10.0 m t =.0 s 1 d = 5.0 m [E] = 5.0 m t = 1.5 s d 3 = 30.0 m [W] = 30.0 m t = 5.0 s 3 average velocity ( v ave ) The total displacement is: d = 10.0 m m + ( 30.0 m) = 15.0 m The total time is: t =.0 s s s = 8.5 s d vave t 15.0 m 8.5 s 1.8 m/s 1.8 m/s [W] The person s average velocity is 1.8 m/s [W]. 16 Copyright 009 Pearson Education Canada

17 . Given Consider forward to be positive. d A = 100 m [forward] = +100 m t = 9.84 s A d B = 00 m [forward] = +00 m t = 19.3 s B d C = 400 m [forward] = +400 m t C = 1.90 min average velocity ( v ave ) The total displacement is: d = +100 m + (+ 00 m) + (+400 m) = +700 m The total time is: 60 s t = 9.84 s s min 1 min = 9.84 s s s = s d vave t 700 m s 4.89 m/s 4.89 m/s [forward] d va t 100 m 9.84 s 10. m/s 10. m/s [forward] d vb t 00 m 19.3 s 10.4 m/s 10.4 m/s [forward] d vc t 400 m s 3.51 m/s 3.51 m/s [forward] The average velocity of all three runners is 4.89 m/s [forward]. Person A s average velocity is 10. m/s [forward] (faster than the average velocity). Person B s average 17 Copyright 009 Pearson Education Canada

18 velocity is 10.4 m/s [forward] (faster than the average velocity). Person C s average velocity is 3.51 m/s [forward] (slower than the average velocity). Student Book page 38 Example 1.8 Practice Problems 1. (a) 3 m/s for s, rest for 3 s, 3 m/s for 4 s, 6 m/s for 1 s (c) d = df -di =-1 m -0 m =-1 m (d) The object is stopped when the velocity-time graph is along the time axis, between 5 s. Example 1.9 Practice Problems 1. (a) Student Book page 40 Displacement is the area under the velocity-time graph. Consider north to be positive. 18 Copyright 009 Pearson Education Canada

19 0 s: 1 A bh 1 d ( s)(+6 m/s) +6 m 6 m [N] 5 s: A lw d (+6 m/s)(3 s) +18 m 18 m [N] 5 7 s: 1 A bh 1 d ( s)(+6 m/s) +6 m 6 m [N] 7 9 s: 1 A bh 1 d ( s)( 6 m/s) 6 m 6 m [S] or 6 m [N] 9 10 s: 1 Alw bh 1 d ( 6 m/s)(10 s 9 s) (10 s 9 s)( 1 m/s ( 6 m/s)) 6 m ( 3 m) 9 m 9 m [S] (b) d or 9 m [N] = +6 m + (+18 m) + (+6 m) + ( 6 m) + ( 9 m) = +15 m = 15 m [N] d (c) vave t 15 m 10 s 1.5 m/s 1.5 m/s [N] 19 Copyright 009 Pearson Education Canada

20 Example 1.10 Practice Problems 1. (a) Student Book page 4 (b) Student Book page 43 Example 1.11 Practice Problems 1. (a) The object travels with uniform motion, changes direction at 10 s, and travels with uniform motion. The object travels forward at 5 m/s for 10 s, then backward at 5 m/s for 8 s. Consider forward to be positive. d = d1+ d = v1t1+ vt æ mö = s ç çè s ( ) æ mö s ø ç çè s ( ) ø =+ 50 m + (-40 m) =+ 10 m = 10 m [forward] (b) 0 Copyright 009 Pearson Education Canada

21 Student Book page Concept Check (a) (b) (c) (d) 1.4 Check and Reflect Knowledge 1. Spaces between dots on a ticker tape for uniform motion are equal. On a ticker tape for accelerated motion, the spaces between dots are unequal or different for equal time intervals. (One would have to do some measuring and calculating to determine if the object is uniformly accelerated.). For an object undergoing uniform motion, the object experiences equal displacement during equal time intervals and its velocity remains constant. For an object undergoing uniformly accelerated motion, the object experiences an equal change in velocity in equal time intervals. 3. The slope of a position-time graph gives velocity. 4. A position-time graph for an object undergoing uniform motion is a straight line (linear): horizontal for objects at rest, and with a positive or negative slope for objects mong at a constant rate. Accelerated motion is represented by a curve (section of a parabola) on a position-time graph. 5. The slope of a velocity-time graph gives acceleration. 6. The object would be undergoing uniform acceleration. 7. Consider east to be positive. D d = area under graph 1 = lw + bh æ m ö = 5.0 [E] 10.0 s ç çè s ( ) 1 æ m ö + ( 5.0 s ) 10.0 [E] ø ç çè s ø = 50 m [E] + 5 m [E] = 75 m [E] 1 Copyright 009 Pearson Education Canada

22 8. Consider up to be positive. d = v1t1+ vt + v3t3 = ( km/ s )(5.0 s s ) ++ (.0 km/s)(7.0 s s ) ++ ( 4.0 km/ s )(10.0 s s ) =+ ( 0 km) ++ ( 4.0 km) ++ ( 1 km) =+ 36 km = 36 km [up] 9. The velocity-time graph for an object undergoing negative acceleration is a line sloping down to the right above or below the time axis. It represents a slowing down motion above the time axis and a speeding up motion (in the negative direction) below the time axis. 10. The area under a velocity-time graph gives displacement. 11. Since an object undergoing uniform motion does not experience a change in velocity, the velocity-time graph is a horizontal line (slope of zero) either at zero (object at rest), or above or below the time axis. A velocity-time graph for an object undergoing uniformly accelerated motion is a line with a positive or negative slope. 1. Assuming the forward direction is positive, an acceleration-time graph for an object that is slowing down in the forward direction is a horizontal line below the time axis. 13. Consider east to be positive. v a = t m/s -(-50 m/s) = 10.0 s m/s = 10.0 s =+ 15 m/s = 15 m/s [E] 14. The acceleration-time graph is a horizontal line at- 1 m/s [N]. Applications 15. Consider west to be positive. 1 d = ( f + i) t + 00 m = 1 ( m/s + ( m/s)) t + 00 m = ( m/s) t t = 8.00 s Copyright 009 Pearson Education Canada

23 - a = t m/s -( m/s) = 8.00 s =+ 1.5 m/s = 1.5 m/s [W] 16. (a) Given Consider north to be positive. v1 70 km/h [N] +70 km/h Δ t1= 75 min v 90 km/h [N] 90 km/h t 96 min average velocity ( v ave ) Average velocity is a vector quantity. Determine the total displacement vector. d = d1+ d = v1t1+ vt æ km 1 h = min ö æ km 1 h ö ++ ç min çè h 60 min ø ç çè h 60 min ø =+ 3 km Determine the total time. t t1 t 1 h min 60 min.85 h Calculate the average velocity. d vave = t + 3 km =.85 h =+ 81 km/h = 81 km/h [N] The average velocity is 81 km/h [N]. (b) 3 Copyright 009 Pearson Education Canada

24 d = d1+ d = lw + lw = v t + v t 1 1 æ km 1 h ö = 70 [N] 75 min æ km 1 h ö + ç 90 [N] 96 min çè h 60 min ø ç çè h 60 min ø = 3 km [N] The total time is 1 h 171 min.85 h 60 min Therefore, the average velocity is d vave = t 3 km [N] =.85 h = 81 km/h [N] 17. Acceleration is the slope of the velocity-time graph. Consider right to be positive. v a = t m/s -( +.0 m/s) = 30.0 s = m/s = 0.33 m/s [right] Copyright 009 Pearson Education Canada

25 Extension s to.0 s: sharp acceleration [E].0 s to 6.0 s: gentle acceleration [E] 6.0 s to 10.0 s: uniform motion [E] 10.0 s to 11.0 s: sharp deceleration, to momentary stop 11.0 s to 1.0 s: sharp acceleration [W] 1.0 s to 15.0 s: medium acceleration [W] 15.0 s to 18.0 s: gentle deceleration [E] 18.0 s to 4.0 s: uniform motion [W] 4.0 s to 7.0 s: medium acceleration [E] to momentary stop 7.0 s to 30.0 s: medium acceleration [E] Student Book page 46 Concept Check Graph Type Reading the Graph Slope Area position-time position velocity velocity-time velocity acceleration displacement acceleration-time acceleration jerk change in velocity Student Book page 47 Example 1.1 Practice Problems 1. Given Consider east to be positive. v i = 6.0 m/s [E] = +6.0 m/s a = 4.0 m/s [E] = +4.0 m/s v f = 36.0 m/s [E] = m/s time ( t ) v Rearrange the equation a. Since you are diding by a vector, use the scalar t form of the equation. v t a 36.0 m/s 6.0 m/s 4.0 m/s 7.5 s It will take the motorcycle 7.5 s to reach a final velocity of 36.0 m/s [E]. 5 Copyright 009 Pearson Education Canada

26 . Given Consider north to be positive. v km 1 h 1000 m i = 0 km/h [N] = m/s h 3600 s 1 km a = 1.5 m/s [N] = +1.5 m/s t = 9.3 s maximum velocity ( v f ) v Use the equation a. t a t at ( 1.5 m/s )(9.3 s) ( 5.6 m/s) m 1 km 3600 s s 1000 m 1 h 70 km/h 70 km/h [N] The maximum velocity of the elk is 70 km/h [N]. Example 1.13 Practice Problems 1. Given Consider south to be positive. v i = 16 m/s [S] = +16 m/s v f = 4.0 m/s [S] = +4.0 m/s t = 4.0 s displacement ( d ) Use the equation 1 d ( v ). i t 1 d ( 16 m/s ( 4.0 m/s))(4.0 s) 40 m 40 m [S] The hound s displacement is 40 m [S]. Student Book page 48 6 Copyright 009 Pearson Education Canada

27 . Given Consider uphill to be positive. v i = 3.0 m/s [uphill] = +3.0 m/s v f = 9.0 m/s [downhill] = 9.0 m/s t = 4.0 s displacement ( d ) Use the equation 1 d ( v ). i t 1 d ( 3.0 m/s ( 9.0 m/s))(4.0 s) 1 m The ball s displacement is 1 m or 1 m [downhill]. Student Book page 50 Example 1.14 Practice Problems 1. Given Consider down to be positive. v i = 3.0 m/s [down] = +3.0 m/s a = 4.0 m/s [down] = +4.0 m/s t = 5.0 s displacement ( d ) 1 Use the equation d t a ( t ). 1 d ( 3.0 m/s)(5.0 s) ( 4.0 m/s )(5.0 s) 65 m 65 m [down] The skier s displacement after 5.0 s is 65 m [down].. Given Consider forward to be positive. v km 1 h 1000 m i = m/s h 3600 s 1 km a = 0.80 m/s t = 1.0 min = 60 s distance ( d ) 7 Copyright 009 Pearson Education Canada

28 1 Use the equation d t a ( t ). 1 d ( 7.8 m/s)(60 s) ( 0.80 m/s )(60 s) 7 m d.310 m The motorcycle travels.3 10 m. Student Book page 51 Example 1.15 Practice Problems 1. Given Consider forward to be positive. d = +150 m v f = 0 a = 15 m/s time ( t ) 1 Use the equation d t a ( t ) to solve for time m (0)( t) ( 15 m/s )( t) 150 m t 7.5 m/s 4.5 s The plane stops after 4.5 s.. Given Consider north to be positive. t = 6. s v f 1 h 1000 m = 160 km/h [N] 44.4 m/s 3600 s 1 km d = 0 m [N] = +0 m acceleration ( a ) 1 Use the equation d t a ( t ). 8 Copyright 009 Pearson Education Canada

29 ( t d) a ( t) (( 44.4 m/s)(6. s) ( 0 m)) (6. s).9 m/s.9 m/s [N] The Corvette s acceleration is.9 m/s [N]. Student Book page 5 Example 1.16 Practice Problems 1. Given 70 m/s [forward] t 9 s 0 m/s (a) acceleration ( a ) (b) distance ( d ) - (a) Use the equation a =. Since the jetliner comes to a halt, final velocity is t zero. - a = t 0 m/s- 70 m/s [forward] = 9 s =-.41 m/s [forward] (b) Use the equation a d. ad d a (0 m/s) (70 m/s) (.41 m/s ) 1015 m 1.0 km (a) The jet s acceleration is.4 m/s [forward]. (b) The runway length is 1.0 km. 9 Copyright 009 Pearson Education Canada

30 . Given 50 km/h 100 km/h a 3.8 m/s distance ( d ) Convert initial and final velocities to m/s. Then use the equation a d to find the on-ramp length. km 1000 m 1 h m/s h 1 km 3600 s km 1000 m 1 h m/s h 1 km 3600 s = + ad - d = a (7.78 m/s) - (13.89 m/s) = (3.8 m/s ) = 76 m The minimum length of the on-ramp is 76 m. Student Book page Check and Reflect Applications 1. Given Consider forward to be positive. D t = 3.0 s a = 1.0 cm/s [forward] =+ 1.0 cm/s = 5.0 cm/s [forward] =+ 5.0 cm/s displacement ( Dd ) 1 Use the equation d v it a t. 1 D d = ( cm/s)(3.0 s) + ( cm/s )(3.0 s) =+ 0 cm = 0 cm [forward] The robot will travel 0 cm [forward]. 30 Copyright 009 Pearson Education Canada

31 . Given Consider right to be positive. 10 m/s [right] 10 m/s 0 m/s [right] 0 m/s t 5.0 s displacement ( Dd ) 1 Since you do not know the acceleration, use the equation D d = ( + ) Dt. 1 D d = ( + 10 m/s + ( + 0 m/s))(5.0 s) =+ 75 m = 75 m [right] The logging truck moves 75 m [right]. 3. Given 0 m/s a 3.75 m/s t 5.65 s distance ( d ) Apply the equation d v 1 it a t, where v i 0. 1 d 0 (3.75 m/s )(5.65 s) 59.9 m The car will travel 59.9 m. 4. Given Consider forward to be positive. 0 m/s 3 t.7510 s 460 m/s [forward] 460 m/s acceleration ( a ) - Use the equation a =. t m/s -0 m/s a = s 5 = m/s 5 = m/s [forward] 31 Copyright 009 Pearson Education Canada

32 5 The bullet s acceleration is m/s [forward]. 5. Given a 4.5 m/s 0 m/s d.6 km = 600 m time ( D t ) Apply the equation d v 1 it a t, where v i 0. 1 d at d t a (600 m ) m 4.5 s 11 s It takes the aircraft 11 s to travel down the runway. 6. Given = 14.0 m/s = 0 m/s D t = 5.60 s distance ( D d ) 1 Since acceleration is uniform, use the equation D d = ( + ) D t. 1 D d = ( 14.0 m/s + 0 m/s )( 5.60 s ) = 39. m The skidding distance is 39. m. 7. Given d 150 m 50 km/h 30 km/h magnitude of acceleration (a) Convert initial and final speeds to m/s. Use the equation a d. 3 Copyright 009 Pearson Education Canada

33 30 50 km 1000 m 1 h 8.33 m/s h 1 km 3600 s km 1000 m 1 h m/s h 1 km 3600 s f i v v ad a d (8.33 m/s) (13.89 m/s) (150 m) 0.41 m/s The magnitude of the car s acceleration is 0.41 m/s. 8. Given d = 1.3 km [forward] = 1300 m [forward] = 90 km/h [forward] = 0 m/s acceleration ( a ) Convert km/h to m/s. km 1000 m 1 h 90 5 m/s h 1 km 3600 s Then use the equation a d. f v a d 0 (5 m/s) (1300 m) 0.4 m/s The train s acceleration is m/s [forward]. 9. Given Consider west to be positive. = 0 m/s D t =.00 s D d = 150 m [W] =+ 150 m acceleration ( a ) 1 Use the equation d t a t, where i v = Copyright 009 Pearson Education Canada

34 ( d - t) a = ( t) ( m-0) = (.00 s) = m/s = 75.0 m/s [W] The rocket s acceleration is 75.0 m/s [W]. 10. Given = 0 m/s = 41 km/h D d = 96.0 m magnitude of acceleration (a) Convert km/h to m/s. km 1000 m 1 h m/s h 1 km 3600 s Use the equation v v ad, where v i 0. f f i v a d (66.94 m/s) 0 m/s (96.0 m) 3.3 m/s The jet s acceleration has a magnitude of 3.3 m/s. 11. Given Consider south to be positive. = 0 m/s d = m [S] = m t = 14.1 s acceleration ( a ) 1 d t a t, where v i 0. Apply the equation 34 Copyright 009 Pearson Education Canada

35 1 d = a( t) d a = ( t) ( m) = (14.1 s) =+ 3.5 m/s = 3.5 m/s [S] The motorcycle s acceleration is 3.5 m/s [S]. 1. Given d 39.0 m 0 m/s 97.0 km/h magnitude of acceleration (a) Use the equation a d, where 0. km 1000 m 1 h m/s h 1 km 3600 s f i v v ad a d 0 (6.94 m/s) (39.0 m) 9.31 m/s The magnitude of the car s acceleration is 9.31 m/s. 13. Given a = 49 m/s [forward] v i = 110 km 1000 m 1 h m/s h 1 km 3600 s v f = 0 distance ( d ) Use the equation a d. 35 Copyright 009 Pearson Education Canada

36 f v d a 0 (30.56 m/s) ( 49 m/s ) 9.5 m The minimum stopping distance must be 9.5 m. 14. Given Consider north to be positive. = 9.0 m/s [N] =+ 9.0 m/s D d = 1.54 km [N] = m D t =.0 min = 10 s acceleration ( a ) 1 Rearrange the equation d v it a t to solve for acceleration. ( d - t) a = ( t) æ æ mö m s ç çè s ( ) ö ç çè ø ø = (10 s) = m/s = m/s [N] The submarine s acceleration is m/s [N]. Example 1.17 Practice Problems 1. Given Consider down to be positive. t = s d = 4.80 m [down] = m a = 9.81 m/s [down] = m/s initial velocity ( v i ) 1 Use the equation d = v ( ) it+ a t. Student Book page Copyright 009 Pearson Education Canada

37 1 d - a( t ) v i = t 1æ mö m - ç ( s) çè s ø = s =+.7 m/s =.7 m/s [down] The rock s initial velocity is.7 m/s [down].. Given Consider down to be positive..0 m/s [down].0 m/s d 1.75 m [down] 1.75 m a 9.81 m/s [down] 9.81 m/s time ( t ) Determine the final velocity using the equation a d. v v ad f i m m (1.75 m) s s 6.19 m/s v Determine the time using a. Use the scalar form of the equation because you are t diding by a vector. v a t t a 6.19 m/s (.0 m/s) 9.81 m/s 0.43 s The football is in the air for 0.43 s. 3. Given Choose down to be positive. a.00 m/s [up].00 m/s 4.00 m/s [down] 4.00 m/s t 1.80 s final velocity ( v f ) distance ( d ) 37 Copyright 009 Pearson Education Canada

38 Use the equation a t v v at f i to find final velocity. m m (1.80 s ) s s m/s = m/s [down] 1 Use the equation d v it a t to find the distance ( d). m 1 m d 4.00 (1.80 s ).00 (1.80 s) s s 3.96 m d 3.96 m The final velocity is m/s [down] and the elevator travels 3.96 m. Student Book page 59 Example 1.18 Practice Problem 1. Given Choose down to be positive. d = 7 m [down] = +7 m a 9.81 m/s [down] 9.81 m/s 0 m/s final speed ( v f ) Determine the final speed using the equation v v a d. f i v v ad f i m (7 m) s 3 m/s The final speed of the riders before they start slowing down is 3 m/s. 38 Copyright 009 Pearson Education Canada

39 Student Book page 60 Example 1.19 Practice Problems 1. (a) Given Choose down to be positive. d = 0.0 m [down] = +0.0 m a 9.81 m/s [down] 9.81 m/s 0 m/s final velocity ( v f ) Determine the final velocity using the equation a d. f i v v ad m (0.0 m) s m/s 19.8 m/s The pebble hits the ground with a velocity of 19.8 m/s [down]. (b) Given Choose down to be positive. d = 0.0 m [down] = +0.0 m a 9.81 m/s [down] 9.81 m/s 0 m/s v f = 19.8 m/s [down] time ( t ) v Determine the time using a. Use the scalar form of the equation because you t are diding by a vector. v a t t a m/s 0 m/s 9.81 m/s.0 s The pebble hits the ground after.0 s. 39 Copyright 009 Pearson Education Canada

40 Student Book page 61 Concept Check (a) The speed of an object at the launch level is the same at the moment it is thrown upwards and at the moment it arrives back to this level. Since an object thrown upward experiences uniformly accelerated motion due to graty, it undergoes equal changes in velocity over equal time intervals according to Dv the equation a =. It therefore makes sense that the time taken to reach maximum D t height is the same time required to fall back down to the launch level. 1 (b) According to the equation d v it a t, the length of time a projectile is in the air depends on the acceleration due to graty and on the initial velocity of the object. This answer is surprising because it is counterintuitive: It is a common misconception that an object s mass affects how long it spends in the air. Student Book page 6 Concept Check The slope of the velocity-time graph for vertical projectile motion should be 9.81 m/s, where [up] is positive. Student Book page Check and Reflect Knowledge 1. A projectile is any object thrown, dropped, or fired into the air.. The height ( d) from which an object is dropped determines how long it will take to reach the ground. Applications 3. Given Choose down to be positive. t s 0 a 9.81 m/s [down] 9.81 m/s height ( d ) 1 Use the equation d v it a t, where v i = 0. d 1 0 ( 9.81 m/s )(3.838 s) 7.3 m 7.3 m [down] The muffin falls from a height of 7.3 m. 40 Copyright 009 Pearson Education Canada

41 4. Given Choose down to be positive. 0 m/s t 0.56 s a 9.81 m/s [down] 9.81 m/s height ( d ) 1 Use the equation d v it a t, where v i = ( 9.81 m/s )(0.56 s) 1.5 m 1.5 m [down] The toys are being dropped from a height of 1.5 m. 5. Given Consider down to be positive. t s d.00 m [down].00 m 0 m/s acceleration ( a ) 1 Use the equation d v it a t, where v i 0. 1 d at d a t.00 m s 1.61 m/s 1.61 m/s [down] The acceleration due to graty on the Moon is 1.61 m/s [down]. 6. Given Consider down to be positive. 5.0 m/s [up] 5.0 m/s di 1.50 m [up] 1.50 m 0 m/s a 9.81 m/s [down] 9.81 m/s 41 Copyright 009 Pearson Education Canada

42 maximum height ( d ) At the instant the ball reaches maximum height, its velocity is zero. The maximum height reached by the ball is the height it reaches from the initial impulse plus the height from which it was launched. Use the equation a d and substitute scalar quantities. ad d a 0 ( 5.0 m/s) (9.81 m/s ) 1.7 m The ball s height from the floor is 1.50 m 1.7 m.8 m. The basketball reaches a maximum height of.8 m. 7. Given Choose down to be positive. v i = 0 m/s t.6 s a 9.81 m/s [down] = m/s final velocity ( v f ) distance ( d ) Because the student drops from rest, initial velocity is zero. For final velocity, use the equation a. t = + at = 0 m/s + ( m/s )(.6 s) =+ 5.5 m/s = 5.5 m/s [down] 1 For distance, use the equation d t. 1 d t 1 (0 m/s 5.5 m/s)(.6 s) 33 m The student falls 33 m and the student s final velocity is 6 m/s [down]. 4 Copyright 009 Pearson Education Canada

43 8. Given Choose down to be positive. d 1.75 m [down] 1.75 m a 4.8 m/s [down] 4.8 m/s v i = 0 time ( t ) 1 Use the equation d v it a t, where v i = 0. Use the scalar form of the equation because you are diding by a vector. d 1 at t d a (1.75 m ) m 4.8 s s The tennis ball takes s to drop 1.75 m on Jupiter. 9. Given Choose down to be positive. t 6.5 s a 9.81 m/s [down] 9.81 m/s distance ( d ) The ball takes 6.5 s 3.15 s to reach its maximum height. At maximum height, v 1 i = 0. To find maximum height, use the equation d v it a t. 1 d 0 (9.81 m/s )(3.15 s) 47.9 m The baseball reaches a maximum height of 47.9 m. 10. Given Choose down to be positive. d 3.0 m [down] 3.0 m a 9.81 m/s [down] 9.81 m/s time ( t ) 43 Copyright 009 Pearson Education Canada

44 The kangaroo s time in the air is t. Find time using the equation 1 d v it a t, where v i = 0 from maximum height. Use the scalar form of the equation because you are diding by a vector. d 1 at t d a (3.0 m ) m 9.81 s 0.78 s Since this time is the time taken to fall from maximum height, the kangaroo s total time in the air is 0.78 s 1.6 s. The kangaroo is in the air for 1.6 s. 11. Given Choose down to be positive. d 190 m [down] 190 m a 9.81 m/s [down] 9.81 m/s 0 m/s time ( t ) 1 Find the time using the equation d v it a t, where v i = 0 from maximum height. Use the scalar form of the equation because you are diding by a vector. 1 ( ) d a t d t a (190 m ) m 9.81 s 6. s The penny takes 6. s to fall 190 m. 1. Given Choose down to be positive. t.75 s di 1.30 m [up] 1.30 m a 9.81 m/s [down] 9.81 m/s 44 Copyright 009 Pearson Education Canada

45 maximum height ( d ) The ball takes.75 s s to reach its maximum height. Find height using the 1 equation d v it a t, where v i = 0 from maximum height. 1 m d (1.375 s) s 9.7 m Total height, from ground, is 9.7 m m = m = 10.6 m The coin reaches a maximum height of 10.6 m. 13. Given Choose down to be positive. d 10 m [down] 10 m a 9.81 m/s [down] 9.81 m/s 0 m/s time ( t ) 1 Use the equation d v it a t, where v i = 0. Use the scalar form of the equation because you are diding by a vector. t d a (10 m ) m 9.81 s 1.4 s The diver takes 1.4 s to reach the water s surface. 14. Given hbuilding 5.0 m vwalking.75 m/s hcatch 1.5 m distance ( d ) If the person catches his keys directly below where they are dropped, the amount of time the keys have to fall is the same amount of time the person has to walk to catch 1 them. To find the time taken for the keys to fall, use the equation d v it a t, where v i = Copyright 009 Pearson Education Canada

46 To find t, use the scalar form of the equation because you are diding by a vector. t d a vertical (5.0 m m ) m s 0.87 s To find the distance the person needs to walk to catch his keys, use the equation d v. t d v t horizontal horizontal m.75 (0.87 s ) s.4 m The person is.4 m away. Extensions 15. Given Choose up to be positive. t 50 s 0 00 m/s [up] 00 m/s a 9.81 m/s [down] 9.81 m/s (a) acceleration ( a ) (b) height at which fuel runs out ( dfuel ) (c) explanation for height gain (d) maximum height ( dmax ) (a) To find acceleration, assume the rocket launches from an initial velocity of zero. v Use the equation a. t 00 m/s 0 a 50 s 4.0 m/s 4.0 m/s [up] 1 (b) Determine the distance travelled upward using the equation d v it a t. d 1 fuel 0 (4.0 m/s )(50 s) m (c) As soon as the rocket s fuel runs out, its upward acceleration suddenly changes to the downward acceleration due to graty. At this moment, the rocket is mong upward with a velocity it gained due to the thrust of its engines. This is the initial 46 Copyright 009 Pearson Education Canada

47 velocity for this last leg of the flight without further thrust. The rocket will continue to rise until its upward velocity is reduced to zero by the (downward) acceleration due to graty. At maximum height, f 0 m/s, i 00 m/s [up] 00 m/s Using f i at, t 00 m/s Therefore, t 0.39 s 0 s 9.81 m/s The rocket contines to gain height for 0 s. (d) To determine how far the rocket will travel once it runs out of fuel, use the equation a d, where v f (at maximum height) equals zero. dmax dfuel dtop ad dtop a 0 (00 m/s) ( 9.81 m/s ) 039 m dmax 5000 m 039 m 7039 m m (a) The rocket s acceleration during fuel burning is 4.0 m/s [up]. 3 (b) At the end of 50 s, the rocket has reached a height of m. 3 (d) The rocket s maximum height is m. 16. Given Choose down to be positive. d 60.0 m [down] 60.0 m t s a 9.81 m/s [down] 9.81 m/s initial velocity of second ball ( v ) i Determine how long the first ball takes to hit the ground. Use the equation 1 d v it a t, where v i = 0. Use the scalar form of the equation because you are diding by a vector. 47 Copyright 009 Pearson Education Canada

48 d v 1 it a t d t1 a (60.0 m) m 9.81 s s The second ball has s less to cover the same distance. t s s.647 s Use the equation 1 d v it a t to determine the initial velocity of the second ball. 1 d at v i t 1 m 60.0 m 9.81 (.647 s) s.647 s 9.68 m/s 9.68 m/s [down] The initial velocity of the second ball was 9.68 m/s [down]. Student Book pages Chapter 1 Reew Knowledge 1. Vector quantities have magnitude and direction. They must be added by vector addition special tip-to-tail procedure. Scalar quantities have magnitude only. Scalar quantity: a mass of 10 kg Vector quantity: a displacement of 10 m [N]. Time (s) Position (cm [right]) m [forward] 10 m [forward] 3. (a) v 1.0 m/s [forward] or 1.0 m/s [backward] 10 s 0 s 0 m [right] 0 m [right] (b) v.0 m/min [right] 10 min 0 min 48 Copyright 009 Pearson Education Canada

49 0 m [forward] ( 5 m [forward]) (c) v 1.7 m/s [forward] 15 s 0 s 4. Given t 15.0 min v 30.0 m/s [W] displacement ( d ) d Use the equation v and convert 15.0 min into seconds. t d m 60 s 30.0 [W] 15.0 min s 1 min m [W] 7.0 km [W] The vehicle s displacement is 7.0 km [W]. 5. Given v = 5.0 km/h D d = 3.50 km time ( D t ) d Use the equation v. t d vt d t v 3.50 km 5.0 km/h 60 min 0.70 h 1 h 4 min and Verify The cross-country skier will take 4 min. Check: (0.70 h)(5.0 km/h) = 3.5 km Dd 6. vave = D t 5.0 m m = 0.0 s = 3.75 m/s Dd vave = D t -5.0 m [right] = 0.0 s =-1.5 m/s [right] or 1.5 m/s [left] 49 Copyright 009 Pearson Education Canada

50 D d = 5.0 m [right] + (-50.0 m [right]) =-5.0 m [right] or 5.0 m [left] 7. A person standing still could have the same average velocity as someone running around a circular track if the runner starts and finishes at the same point, ensuring total displacement is zero. Average speed would be different since the stationary person will have no change in distance, while the runner will have covered a certain distance in the same amount of time. 8. If an object is in the air for 5.6 s, it reaches maximum height in half the time, or 5.6 s.8 s. 9. An object is in the air for twice the amount of time it takes to reach maximum height, or 3.5 s 7.0 s, proded it lands at the same height from which it was launched. 10. The initial vertical velocity for an object dropped from rest is zero. Applications 11. Given v = 0.77 m/s D d = 150 m time ( D t ) d Use the equation v. t Dd D t = v 150 m = 0.77 m/s = s The diver takes s to travel 150 m. 1. Given D d = 4 km [W] D t = 8.0 h average velocity ( v ave ) d Use the equation vave. t v 4 km [W] ave = 8.0 h km 1000 m 1 h = 5.5 [W] h 1 km 3600 s =1.5 m/s [W] Terry Fox s average velocity was 1.5 m/s [W]. 50 Copyright 009 Pearson Education Canada

51 13. The point of intersection on a position-time graph shows the time and location where two objects meet. The point of intersection on a velocity-time graph shows when two objects have the same velocity. 14. Given Consider north to be positive. vthief = 5.0 m/s [N] =+ 5.0 m/s dthief = 0 m 1 vofficer = 7.5 m/s [N] =+ 7.5 m/s dofficer = 0 m [S] =-0 m 1 distance ( D dofficer ) From the graph, the police officer catches up with the thief after about 8.0 s. Equation for thief: ya 5.0 t Equation for officer: yb 7.5t 0 At the point of intersection, the time is ya yb 5.0t 7.5t 0 0.5t t 8.0 s Displacement is: D dofficer = dofficer -d officer1 = yb -(-0 m) = ( m/s)(8.0 s) -0 m -(-0 m) =+ 60 m = 60 m [N] and Verify The police officer will run 60 m before catching the thief. Check: Distance run in 8.0 s at 7.5 m/s is 60 m. 15. Given 100 m/s 0 m/s d 1.0 cm m magnitude of acceleration (a) The bullet comes to rest within the vest, so 0. Use the equation v v ad. f i 51 Copyright 009 Pearson Education Canada

52 f i v v ad a d 0 (100 m/s) (0.010 m) m/s 7 The magnitude of the bullet s acceleration is m/s. 16. Given v 35 km/h [forward] t 30min 0.50 h distance ( d ) Determine the area under the graph after 0.50 h. km d 35 [forward] (0.50 h ) h 18 km [forward] The elk will travel 18 km. 17. Given t.50 min v 89 km/h distance ( d ) d Use the equation v. t d = vt km 1 h = min h 60 min = 34.5 km The speedboat will travel 34.5 km. 18. Given d 300 km tstagecoach 4 h tairliner 0 min v airliner speed factor v stagecoach 5 Copyright 009 Pearson Education Canada

53 Determine the average speed of the stagecoach and airliner and then compare the two speeds. Stagecoach: d v t 300 km 4 h 1.5 km/h Airliner: d v t 300 km 0 min 1 h 60 min 900 km/h 900 km/h km/h = The airliner is 7 times faster than the stagecoach. 19. Given di 647 km df 3 09 km t 5.0 h average speed ( v ave ) Determine the distance by subtracting the two odometer readings. Use the equation d v. t d vave t 3 09 km 647 km 5.0 h km/h km 1000 m 1 h h 1 km 3600 s 31 m/s The car s average speed was km/h, which equals 31 m/s. 53 Copyright 009 Pearson Education Canada

54 0. Given Choose downhill to be positive. d 90.0 m [downhill] 90.0 m t 8.00 s 0 m/s acceleration ( a ) final velocity ( v f ) 1 Find final velocity using the equation d t, where 0. d = - t ( m) = s =+.5 m/s =.5 m/s [downhill] v Find acceleration using the equation a. t +.5 m/s -0 m/s a = 8.00 s =+.81 m/s =.81 m/s [downhill] The motorcycle s velocity at 8.00 s is.5 m/s [downhill]. Its acceleration is.81 m/s [downhill]. 1. Given Choose north to be positive. t = 4.0 s d = 30.0 m [N] = m = 5.0 m/s [N] =+ 5.0 m/s acceleration ( a ) final velocity ( v f ) 1 First determine the acceleration of the cyclist using the equation d = v ( ) it+ a t. 54 Copyright 009 Pearson Education Canada

55 ( d - t) a = ( t) é æ mö m s ç çè s ( ) ù ê ø ú = ë û (4.0 s) =+ 1.5 m/s = 1.5 m/s [N] - Calculate final velocity using a =. t = + at m æ mö = s s ç çè s ( ) ø =+ 10 m/s = 10 m/s [N] At the other end of the bridge, the cyclist s acceleration was 1.3 m/s [N] and her final velocity was10 m/s [N].. Given Consider south to be positive m/s [S] 10.0 m/s d 70 m [S] 70 m t 45.0 s average velocity ( v ave ) final velocity ( v f ) acceleration ( a ) Dd Use vave = to find average velocity. D t Dd vave = D t + 70 m = 45.0 s = m/s = 16.0 m/s [S] ( + ) Use D vave = to find final velocity. = Dvave - = ( m/s) -( m/s) =+.0 m/s =.0 m/s [S] 55 Copyright 009 Pearson Education Canada

56 - Use a = to find acceleration. D t - a = D t +.0 m/s -( m/s) = 45.0 s = m/s = 0.67 m/s [S] The object has an average velocity of 16.0 m/s [S], a final velocity of.0 m/s [S], and an acceleration of 0.67 m/s [S]. 3. Given t1 t 65 s.88 km d km 1440 m d 1440 m v1 60 m/s speed ( v ) Determine the amount of time the car takes to complete the first half-lap using the Dd equation v =. D t d1 t1 v m 60 m/s 4 s Determine the speed for the second half-lap by diding the distance by the remaining time. d v t 1440 m 65 s 4 s 35 m/s The car s speed for the second half-lap must be 35 m/s. 4. Given vcar = 19.4 m/s v = 0 m/s a police i police = 3. m/s 56 Copyright 009 Pearson Education Canada

57 time ( D t ) final speed ( v police ) f Use the given data to write motion equations for the cars. equation for car: ya 19.4 t equation for police car: yb 1 3. t yb 1.6t At the point of intersection, time is ya yb 19.4t 1.6t t 0 s or 1.1 s Take the second solution: 1.1 s - Use the equation a = to find the final speed of the police car. D t at m s s 39 m/s It takes the police car 1 s to catch the motorist. The police car s final speed is 39 m/s. The scenario is likely to happen on a straight part of a highway, and less likely in a high-speed chase within city limits. 5. Assume both cars start at position 0.0 m [N] and time 0.0 s. The cars pass one another at time t where they have travelled the same displacement. Determine the displacement by calculating the area under the given velocity-time graph for each car. m For car 1 at t, d1 1.0 [N] t s 1 m m For car at t, d (6.0 s) 18.0 [N] (18.0 [N])( t6.0 s) s s Since d 1= d, it follows that 57 Copyright 009 Pearson Education Canada

58 1 (1.0 m/s) t (6.0 s)(18.0 m/s) (18.0 m/s)( t 6.0 s) (1.0 m/s) t 54 m (18.0 m/s) t 108 m (6.0 m/s) t 54 m t 9.0 s The two cars pass one another at 9.0 s. d 1 = (1.0 m/s [N]) t = 1.0 m/s [N] 9.0 s = 108 m [N] = m [N] The cars pass one another at position m [N]. 6. Choose right to be positive. km d 4.0 [right] 5.0 h h 0 km [right] 0 km [left] a = 0 m/s because the velocity-time graph is a horizontal line. 7. Given Consider forward to be positive. a = 9.85 m/s [forward] = m/s = 0 m/s D d = 40 m [forward] =+ 40 m time ( D t ) Use the equation 1 d v ( ) D = id t+ a Dt, where v i = 0. Use the scalar form of the equation because you are diding by a vector. D d = v 1 ( ) it+ a Dt Dd D t = a (40 m) = 9.85 m/s = 9.03 s The fire truck takes 9.03 s to travel 40 m. 8. Given Consider forward to be positive. 5.0 m/s [forward] 5.0 m/s a 3.75 m/s [forward] 3.75 m/s d 95.0 m [forward] 95.0 m 58 Copyright 009 Pearson Education Canada

59 reaction time ( t ) Find the displacement during braking (less than 95.0 m [forward]) using the equation a d, where v f = 0. Reaction time is time for the motion phase before braking begins. The distance travelled while braking is ad d a 0 (5.0 m/s) ( 3.75 m/s ) m The driver must react while travelling a maximum distance of 95.0 m m = m. Maximum reaction time is: d v t d t v m m 5.0 s s The driver has s to react in order to avoid hitting the obstacle. 9. Given d 1.10 km 1100 m km/h 60.0 km/h magnitude of acceleration (a) Convert km/h to m/s. km 1000 m 1 h m/s h 1 km 3600 s km 1000 m 1 h m/s h 1 km 3600 s Use the equation v v ad to calculate the magnitude of acceleration. f i 59 Copyright 009 Pearson Education Canada