(b) A particle with v > 0 and a < 0 is moving forward with decreasing speed. Example: a car slowing down before exiting an Eastbound freeway.

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1 PHY 302 K. Solutions for Problem set # 2. Non-textbook problem #1: (a) A particle with v > 0 and a > 0 is moving forward (i.e., in the positive direction) with increasing speed. Example: a car entering an Eastbound freeway and accelerating to the traffic speed. (Taking the Eastbound direction to be positive.) (b) A particle with v > 0 and a < 0 is moving forward with decreasing speed. Example: a car slowing down before exiting an Eastbound freeway. (c) A particle with v < 0 and a > 0 is moving backward (i.e., in the negative direction) with decreasing speed. Example: a car on a Westbound freeway slowing down before exiting (positive direction being to the East). (d) A particle with v < 0 and a < 0 is moving backward (i.e., in the negative direction) with increasing speed. Example: a car entering a Westbound freeway and accelerating to the traffic speed (positive direction being to the East). (e) A particle with v > 0 and a 0 is moving forward with a speed that neither increases nor decreases; if a(t) stays equal to zero during some time interval, then the speed stays constant during that interval. Example: a car cruising at constant speed on an Eastbound freeway. (f) A particle with v < 0 and a 0 is moving in the negative direction with a speed that neither increases nor decreases; if a(t) stays equal to zero during some time interval, then the speed stays constant during that interval. Example: a car cruising at constant speed on a Westbound freeway. (g) A particle with v 0 and a > 0 is momentarily at rest but is accelerating in the positive direction. Example: A car at the traffic light facing East (the positive direction); when the driver sees the light turning green and hits the gas pedal, the car has a > 0 but for a moment it still has v 0. (h) A particle with v 0 and a < 0 is momentarily at rest but is accelerating in the negative direction. Example: A car at the traffic light facing West (the negative direction); when 1

2 the driver sees the light turning green and hits the gas pedal, the car has a < 0 but for a moment it still has v 0. Non-textbook problem #2: To fix a coordinate system, let s put the origin in Austin and take the Northbound direction on I 35 to be positive. Then Austin is at x 0, San Antonio at x 80 miles, and Waco is at x +100 miles. The net displacement of the driver in question is from Austin to San Antonio. x x Austin x SanAntonio ( 90 mi) (0 mi) 80 mi. (1) It does not matter that he drove to Waco first; he could have driven from Austin to Minneapolis before coming to San Antonio, and the net displacement would still be the same from Austin to San Antonio, regardless of the route. Hence, his average velocity was v avg x t 80 mi 4 hr 20 mi/hr. (2) On the other hand, the detour through Waco did affect the net distance of the trip in question. First, the driver went 100 miles from Austin to Waco, then another 100 miles from Waco back to Austin, and finally 80 more miles from Austin to San Antonio, giving the total mileage s 100 mi mi + 80 mi 280 miles. (Compare to net displacement x +100 mi 100 mi 80 mi 80 miles.) Consequently, the average speed of the trip was v + avg s t 280 mi 4 hr 70 mi/hr. Note that the speed is always positive but the velocity can be either positive or negative, depending on the direction of motion. Likewise, the average speed is always positive, but the average velocity can be either positive or negative, and the sign depends on the direction of the net displacement and also on which direction we consider positive. So if you ve chosen the positive direction to be Southbound instead of Northbound, then you should get average velocity +20 mi/hr instead of 20 mi/hr. But in any case, the average speed should be +70 mi/hr. 2

3 Textbook problem 2.6: Given the positions of the particle at times t 1 and t 2, we can immediately find its net displacement x x 2 x cm 3.4 cm +5.1 cm. (3) Note that this net displacement does not depend on how the particle got from point x 1 to point x 2 it could have gone around the Moon for all we care, the net displacement depends only on the initial and final points. The displacement (3) happened during the time interval t t 2 t 1 (+4.5 s) ( 2.0 s) 6.5 s. (4) Therefore, the average velocity during this interval was v avg x t +5.1 cm 6.5 s cm/s. (5) Note that this is average velocity of the particle rather than its average speed. To find the average speed, we would need to know the net distance L traveled by the particle rather than its net displacement x. But we don t know how the particle got from point x 1 to point x 2 : maybe it was going straight, always moving in the same direction, or maybe first went from x 1 to x cm and then back to x 2, or maybe it was oscillating back and forth between x 1 and x 2 for 137 times. All we know is the net displacement, so we can be sure L x 5.1 cm, but we cannot tell if L is equal or larger than 5.1 cm. Consequently, we do not know the particle s average speed all we have is a lower bound v avg v avg 0.87 cm/s, (6) but we don t know if the average speed is equal or larger than this bound. 3

4 Textbook problem 2.7: (a) This is a tricky question, so let s first write down equations for all the variables, and then we can figure out how to find the unknown variables in terms of what we are given in this problem. The drive home happens in two parts, first fast then slow, and during each part, the time and the distance are related by known speeds, L 1 v 1 t 1, v 1 95 km/h, (7) L 2 v 2 t 2, v 2 65 km/h. (8) The distances traveled during the two stages add up to the net distance, and likewise for the times, L 1 + L 2 L net, (9) t 1 + t 2 t net. (10) We are given L km and v 1 95 km/h, which allows us to find the time duration of the first part of the drive, t 1 L 1 v km 95 km/h 1.37 h 1 h 22 min. (11) For the second part, we are not given either time or distance, but we know the net time of the whole drive t net 3 h 20 min. Subtracting the time of the first part, we find the time of the second t 2 t net t h 1.37 h 1.96 h 1 h 58 min. (12) Consequently, the distance traveled during the second part is L 2 v 2 t 2 65 km/h 1.96 h 128 km, (13) and the net distance is L net L 1 + L km km 258 km 260 km. (14) (b) At this point, we know both the net distance traveled and the net time, so the average 4

5 speed is v avg L net t net 258 km 3.33 h 77 km/h. (15) Textbook problem 2.12: Let s write down equations of motion for the car and the truck, x c (t) x c 0 + v c t, x t (t) x t 0 + v t t, (16) where the superscripts c or t distinguish positions and velocity of the car from those of the truck. The car catches up with the truck at the times t cu when x c (t cu ) x t (t cu ); combining this condition with the equations of motion (16), we get x c 0 + v c t cu x t 0 + v t t cu (17) and hence t cu xt 0 xc 0 v c v t. (18) Note the difference x t 0 xc 0 in the numerator of this formula. This difference is the initial distance between the car and the truck, which we know was 110 meters or 0.11 km. Thus, the time the car needs to catch up with the truck is t cu xt 0 xc 0 v c v t 0.11 km (88 km/h) (75 km/h) 0.11 km 13 km/h h 30 s. (19) Textbook problem 2.18: The acceleration is the rate at which velocity changes with time. Given the acceleration a 1.6 m/s 2, the time the car need to change its velocity by v 110 km/h 80 km/h 30 km/h 8.3 m/s (20) is t δv a 8.3 m/s 5.2 s. (21) 1.6 m/s2 Note that since the acceleration a is given in units of m/s 2, it s important to convert the velocity change v into m/s, otherwise the ratio v/a would give you time not in seconds 5

6 but in rather weird units of t 30 km/h 1.6 m/s 2 19 (km s2 )/(m h). (22) Despite the weirdness, these are units of time and it s possible to convert them to seconds and get 5.2 s as in eq. (21). But it s much easier to first convert v to the m/s units and only then plug it into the t v/a formula. Textbook problem 2.23: The plane accelerates at constant rate a starting from zero velocity v 0 0. At time t from the beginning of acceleration, the plane s velocity is v(t) v 0 + at at (23) and its displacement is x(t) x(t) x 0 v 0 t at2 1 2 at2. (24) Consequently, the time t to the plane needs to accelerate to take-off velocity v to 33 m/s 75 mi/h is t to v to a 33 m/s 11 s, (25) 3.0 m/s2 and during this time the planes moves through distance x to 2 1at2 to a ( 2 vto ) 2 v to 2 a 2a (33 m/s)2 182 m 600 ft. (26) m/s2 So the runway must be at least 600 ft long, or this plane would not be able to takeoff. 6

7 Textbook problem 2.28: How much distance does it take to stop a car moving with initial velocity v 0? First, the driver needs to notice the danger and push the break pedal. This takes a finite reaction time t r during which the car moves at constant velocity v 0 and covers distance L r v 0 t r. (27) Once the breaks engage, the car start decelerating at approximately constant rate and its velocity decreases according to v(t) v 0 + at v 0 a t. (28) Note negative acceleration a a. Eventually, car s velocity v drops to zero at time t b v 0 a (29) and the car stops. While the car is decelerating, its position depends on time according to x(t) x 0 + v 0 t a t2 x 0 + v 0 t 1 2 a t2, (30) so the net displacement of the car while it breaks to stop is L b x(t b ) x 0 v 0 t b 1 2 a t2 b v 0 v 0 a a ( ) 2 2 v0 v2 0 a 2 a. (31) Altogether, the net stopping distance is L stop L r + L b v 0 t r + v2 0 2 a. (32) Now suppose the car moves with initial velocity v 0 95 km/h 59 mi/h 26.4 m/s, the road is dry and the car has good tires so it can decelerate at a 8.0 m/s 2, but the driver s reaction time is a whole second, t r 1.0 s. (The driver must be drunk, or very tires, 7

8 or doing something he should not; a sober alert driver has reaction time between 0.2 s and 0.3 s.) Under these assumptions, the stopping distance is L stop 26.4 m/s 1.0 s + (26.4 m/s)2 70 m 230 ft. (33) m/s2 If the car s deceleration rate is limited to only a 4.0 m/s 2 because the tires are bad or the road is wet and slippery, then the stopping distance increases to L stop 26.4 m/s 1.0 s + (26.4 m/s)2 110 m 370 ft. (34) m/s2 Textbook problem 2.35: To solve this problem we need to make a few assumptions. First, we assume King Kong was so massive that the air resistance didn t slow down his fall. This allows us to use free-fall equations a y (t) g const, v y (t) v 0 gt, (35) y(t) y 0 + v 0 t 1 2 gt2, where y is the vertical coordinate (positive direction up, zero at ground level). Second, large as King Kong was, we assume that the Empire State building is much larger, so me may neglect King Kong s own height when we analyze his fall. Consequently, we set his initial position y 0 to the building height, y m and assume he lands on the ground when y(t) 0. Finally, we assume King Kong fell without being pushed up or down, so we take his initial vertical velocity v 0 to be zero. Under these assumptions, King Kong s fall from the Empire State building is governed by a simple equation y(t) y gt2. (36) The time t f it took him to hit the ground follows from the condition y(t f ) y gt2 f 0. (37) 8

9 Solving this equation, we obtain t f 2y0 g m 9.8 m/s s s. (38) During this time, King Kong s vertical velocity has reached v y (t f ) v 0 gt f gt f 9.8 m/s s 86 m/s 310 km/h 190 mi/h, or in terms of speed, v +86 m/s +190 mi/h. (39) Textbook problem 2.50: (a) Rabbit s position x as a function of time t is plotted on figure For the first 20 seconds or so, the x(t) line look quite straight, which indicates approximately constant velocity during this time. Between t 0 s and t 18 s, the rabbit moves from x 0 m to x 5 m, so his average velocity during this time was about (5 m)/(18 s) 0.28 m/s 1 km/h 0.6 mi/h. And since velocity was constant during that time, the instantaneous velocity at time t 10 s was also 0.28 m/s 1km/h 0.6 mi/h. (b) Draw a straight line tangent to the x(t) curve at point t 30 s, x 16 m. instantaneous velocity at time t 30 s is the slope of this tangent. To read the slope of the graph, note that the tangent crosses the t axis at t 18 s, so its slope is x t rmtangent 16 m 0 m 30 s 18 s The 1.3 m/s. (40) Consequently, v(t 30 s) 1.3 m/s 3 mi/h. (c) The velocity stays constant between t 0 s and t 20 s, so for any time interval during this period the average velocity is the same as in part (a), namely v avg 0.28 m/s 1km/h 0.6 mi/h. (d) At t 25 s, x 8 m, while at t 30 s, x 16 m. This gives us net displacement x 16 m 8 m 8 m during time interval t 30 s 25 s 5 s, and hence average velocity v avg (8 m)/(5 s) 1.6 m/s 6 km/h 3.5 mi/h. 9

10 (e) Finally, between t 40 s and t 50 s, the rabbit moves from x 20 m to x 10 m. The net displacement during this t 10 s period is x 10 m 20 m 10 m, hence negative average velocity v avg ( 10 m)/(10 s) 1 m/s 3.6 km/h 2.2 mi/h. Textbook problem 2.52(a): Looking at figure 2 35 we see that the car was in second gear between t 4 s and t 8s, and during this time the velocity has increased from 13 m/s 29 mi/h to 23 m/s 51 mi/h. Thus, we have net velocity gain v 10 m/s in time t 4 s, hence average acceleration a avg v t 10 m/s 4 s 2.5 m/s 2. (41) Likewise, the car was in the fourth gear between times t 17 s and t 27 s, and during this time the velocity has increased from v 37 m/s 83 mi/hr to v 44 m/s 98 mi/h. This gives us net velocity gain v 7 m/s in time t 10 s and hence average acceleration a avg v t 7 m/s 10 s 0.7 m/s 2. (42) PS: The speeds at which this car changes gears look way too high. Although this could be some very unusual car model, most likely Dr. Giancoli copied this problem without checking from some other book that made a boneheaded unit mistake: Took a real plot where the speeds were in miles per hour, and re-labeled the units as meters per second without converting the numbers. Or perhaps the author of this problem simply made up the numbers without a reality check. 10

Position, Speed and Velocity Position is a variable that gives your location relative to an origin. The origin is the place where position equals 0.

Position, Speed and Velocity Position is a variable that gives your location relative to an origin. The origin is the place where position equals 0. The position of this car at 50 cm describes where the