PUM Physics II - Kinematics Lesson 12 Solutions Page 1 of 16

Transcription

1 PUM Physics II - Kinematics Lesson 12 Solutions Page 1 of Hypothesize (Derive a Mathematical Model) Graphically we know that the area beneath a velocity vs. time graph line represents the displacement of an object. For constant velocity objects we developed the following model: x(t) = x o + v o t When the velocity is changing as it did in the experiment with the falling ball we can find the area under the velocity vs. time graph line with a little geometry. The area of a triangle is: ½ (height)*(base) The base of our triangle is the time interval Δt (which is the same as t, while our height is the change in velocity Δv. ½ ΔvΔt If substitute Δv for an expression with the acceleration a a = Δv/Δt aδt = Δv We get the following expression for the area under the velocity vs. time graph line, which corresponds to the displacement of an accelerating object: Which simplifies to: ½ aδt 2 ½at 2 If we take Δt = t - t i and set t i to 0. If our object has an initial velocity then the y-intercept of the velocity vs. time graph shifts. For example:

2 PUM Physics II - Kinematics Lesson 12 Solutions Page 2 of 16 ½at 2 v o v o t t The displacement then becomes the combined area of the top triangle and the rectangle beneath plus any initial displacement. The generalized equation for displacement then is: This assumes constant acceleration. x(t) = x 0 + v 0 t + ½at Test Your Ideas with Phet Simulations Scenario 1: a) x(t) = (-8 m) + ½(0.75 m/s 2 )t 2 b)

3 PUM Physics II - Kinematics Lesson 12 Solutions Page 3 of 16 c) The functions, written descriptions and graphs are all consistent with each other. The object starts at rest and accelerates at a constant rate resulting in a velocity vs. time graph that should start at zero and have a steady slope. The position vs. time graph on the other hand should start at -8 m (the tree) and increase exponentially, meaning the slope of the line should get steeper and steeper with every second as the velocity gets larger and larger. d) x(t) = (-8 m) + ½(0.75 m/s 2 )t 2 x(t) = (-8 m) + ½(0.75 m/s 2 )(5.0s) 2 x = 1.4 m e) The prediction and outcome agree. Scenario 2: a) x(t) = (-7 m) + (0.75 m/s)t + ½(0.2 m/s 2 )t 2 b)

4 PUM Physics II - Kinematics Lesson 12 Solutions Page 4 of 16 c) Everything is consistent. The position vs. time graph intercepts the y-axis at -7 m and increases exponentially, as expected with accelerated motion. The velocity vs. time graph intercepts the y-axis at 0.75 m/s (his initial velocity) and increases at a constant rate of 0.2m/s 2. d) x(t) = (-7 m) + (0.75 m/s)t + ½(0.2 m/s 2 )t 2 0 m = (-7 m) + (0.75 m/s)t + ½(0.2 m/s 2 )t 2 Quadratic Equation: c + bx + ax 2 c = -7 m b = 0.75 m/s a = 0.2 m/s 2 x = t t = 5.41 s, s Because time cannot be negative: t = 5.41 seconds e) The prediction and outcome agree. Scenario 3: a) x(t) = (5 m) + (-1.0 m/s)t + ½(-0.5 m/s 2 )t 2

5 PUM Physics II - Kinematics Lesson 12 Solutions Page 5 of 16 b) c) Everything is consistent. The position vs. time graph intercepts the y-axis at 5 m and goes towards zero with the slope getting more and more negative, as expected with accelerated motion where the object is speeding up in the negative direction. The velocity vs. time graph intercepts the y-axis at -1.0 m/s (his initial velocity) and decreases at a constant rate of -0.5m/s 2. d) Find the time that the man is moving at 5 m/s: v(t) = v o + at t = (v - v o )/a t = [(-5 m/s) (-1 m/s)]/(-0.5 m/s 2 ) t = 8 s Find the position at 8 s: x(t) = (5 m) + (-1.0 m/s)t + ½(-0.5 m/s 2 )t 2

6 PUM Physics II - Kinematics Lesson 12 Solutions Page 6 of 16 x(t) = (5 m) + (-1.0 m/s)(8 s) + ½(-0.5 m/s 2 )(8 s) 2 x = -19 m e) The prediction and outcome agree. Scenario 4 a) x(t) = (8 m) + (-7.0 m/s)t + (1.0 m/s 2 )t 2 b) c) Everything is consistent. The position vs. time graph intercepts the y-axis at 8 m and goes towards zero with the slope getting less and less steep, as expected with accelerated motion where the object is slowing down in the positive direction. The velocity vs. time graph intercepts the y-axis at -7.0 m/s (his initial velocity) and increases at a constant rate of -1.0 m/s 2, meaning his speed is decreasing at 1.0 m/s 2. d) Find the time that the man is moving at 0 m/s:

7 PUM Physics II - Kinematics Lesson 12 Solutions Page 7 of 16 v(t) = v o + at t = (v - v o )/a t = [(0 m/s) (-7 m/s)]/(1.0 m/s 2 ) t = 7 s Find the position at 7 s: x(t) = (8 m) + (-7.0 m/s)t + (1.0 m/s 2 )t 2 x(t) = (8 m) + (-7.0 m/s)(7.0 s) + (1.0 m/s 2 )(7.0 s) 2 x = m e) The prediction and outcome agree Practice a) The slope of the line segments on the velocity vs. time graph tell me about the acceleration of the motorcycle. It has a positive acceleration from 20 to 65 s, zero acceleration from 0 to 20 s and 65 to 85 s, and negative acceleration from 85 s to 100 s. b) First 20 s: x(t) = (-5 m/s)t x = (-5 m/s)(20 s) = -100 m Distance: 100 m From 20 to 45 s: a = [(0 m/s) (-5 m/s)]/(25 s) = 0.2 m/s 2 x(t) = (-5 m/s)t + ½(0.2 m/s 2 )t 2 x = (-5 m/s)(25 s) + ½(0.2 m/s 2 )(25 s) 2 x = m Distance: 62.5 m From 45 s to 65 s: a = (12 m/s 0 m/s)/(20 s) = 0.6 m/s 2 x(t) = ½(0.6 m/s 2 )t 2 x = ½(0.6 m/s 2 )(20 s) 2 = 120 m Distance: 120 m From 65 s to 85 s: x(t) = (12 m/s)t x = (12 m/s)(20 s) = 240 m Distance = 240 m From 85 s to 100 s: a = (6m/s 12 m/s)/(15 s) = -0.4 m/s 2 x(t) = (12 m/s)t + ½(-0.4 m/s 2 )t 2 x = (12 m/s)(15 s) + ½(-0.4 m/s 2 )(15 s) 2 = 135 m c) Path Length: 100 m m m m m = Displacement: (-100 m) + (-62.5 m) m m m = m

8 PUM Physics II - Kinematics Lesson 12 Solutions Page 8 of Practice Sketch and Translate 1.5 m/s 2 0 Represent Physically Motion Diagram: Reference frame, bus v 1 v 2 v 3 v 4 v 5 Δv 12 Δv 23 Δv 34 Δv 45 0 Starting at rest means the bus had an initial velocity of 0 m/s. Represent Mathematically/Solve and Evaluate x(t) = ½(2.0 m/s 2 )t 2 x = ½(2.0 m/s 2 )(5.0 s) 2 = 25 m 12.5 Practice Sketch and Translate 8 m/s 2 m/s 0 Δv

9 PUM Physics II - Kinematics Lesson 12 Solutions Page 9 of 16 Represent Physically v 4 v 1 v 2 v 3 + Δv 12 Δv 23 Δv 34 We are assuming the bicycle accelerates a constant rate from 8 m/s to 2 m/s. Represent Mathematically/Solve and Evaluate ( v f! vi ) a = " t a = (2 m/s 8 m/s)/(3 s) = -2 m/s 2 x(t) = (8 m/s)t + ½(-2 m/s 2 )t 2 x = (8 m/s)(3 s) + ½(-2 m/s 2 )(3 s) 2 = 15 m 12.6 Reason The car is moving at +14 m/s initially and is accelerating in the positive direction at +3 m/s 2. v 1 v 2 v 3 v 4 v 5 0 Δv 12 Δv 23 Δv 34 Δv 45 3 m/s 2 14 m/s 0 a) v(t) = v o + at v(t) = 14 m/s + (3 m/s 2 )t b) Position after 5 s: x(t) = (14 m/s)t + ½(3 m/s 2 )t 2 x = (14 m/s)(5 s) + ½(3 m/s 2 )(5 s) 2 = m Velocity after 5 s: v(t) = 14 m/s + (3 m/s 2 )t v = 14 m/s + (3 m/s 2 )(5 s) = 29 m/s

10 PUM Physics II - Kinematics Lesson 12 Solutions Page 10 of 16 c) The car starts at -12 m moving at 7 m/s in the positive direction, slowing down at 0.4 m/s per second. v 1 v 2 v 3-12 m Δv 12 Δv m/s 2-12 m a. v(t) = v o + at v(t) = 7 m/s + (-0.4 m/s 2 )t b. Position after 5 s: x(t) = (-12 m) + (7 m/s)t + ½(-0.4 m/s 2 )t 2 x = (-12 m) + (7 m/s)(5 s) + ½(-0.4 m/s 2 )(5 s) 2 = 28 m Velocity after 5 s: v(t) = 7 m/s + (-0.4 m/s 2 )t v = 7 m/s + (-0.4 m/s 2 )(5 s) = 5 m/s d) In class, will vary Compare and Contrast 7 m/s a) A car is pulling out of the driveway moving in the negative direction at a constant velocity. v 4 v 3 v 2 v 1 Δv 12 Δv 23 Δv 34 b) A car is passing another car on the highway, speeding up to get past. v 1 v 2 v 3 v 4 Δv 12 Δv 23 Δv 34 c) A car is slowing down at a red light. v 1 v 2 v 3 v 4 Δv 12 Δv 23 Δv 34

11 PUM Physics II - Kinematics Lesson 12 Solutions Page 11 of 16 d) A car is passing another car on the highway, speeding up to get past. The coordinate system has been changed. v 4 v 3 v 2 v Reason Δv 34 Δv 23 Δv 12 a) The origin is where the car is initially on the left, x o = 0. b) The diagram was missing velocity arrows to indicate the direction of motion. The change in velocity arrow was adjusted to match the motion diagram. Part I Part II Wants to stop: t o = 0 x o = 0 a 0,1 = 0 Foot gets to brake: t 1 = reaction time x 1 = position when foot hits brake a 1,2 Finally stops: t 2 = time when car stops x 2 = final position when car stops c) During Part I, we need to assume that there is no acceleration so that we can predict the distance the car will travel during the time interval needed to react. During Part II we are assuming that the acceleration is constant otherwise our mathematical model for accelerated motion will not apply here and we will be unable to predict t 2 or x 2

12 PUM Physics II - Kinematics Lesson 12 Solutions Page 12 of Represent and Reason Answers will vary depending on coordinate system. a) Sketch 12 m/s -6 m/s 2 0 meters 20 meters b) Motion Diagram v 1 v 2 v 3 v 4 c) Position vs. Time Δv 12 Δv 23 Δv 34

13 PUM Physics II - Kinematics Lesson 12 Solutions Page 13 of 16 d) Velocity vs. Time e) x(t) = (12 m/s)t + ½(-6 m/s 2 )t 2 v(t) = (12 m/s) + (-6 m/s 2 )t f) Time when car stops: 0 m/s = (12 m/s) + (-6 m/s 2 )t t = (-12 m/s)/(-6 m/s 2 ) = 2 s Position when car stops: x = (12 m/s)(2 s) + ½(-6 m/s 2 )(2 s) 2 = 12 m The mystery machine stops in 12 meters, or 8 meters from the stop sign. It takes 2 seconds to come to a complete stop Represent and Reason 26 m/s -3.5 m/s 2 0 meters a) Motion Diagram v 1 v 2 v 3 v 4 Δv 12 Δv 23 Δv 34

14 PUM Physics II - Kinematics Lesson 12 Solutions Page 14 of 16 b) x(t) = (26 m/s)t + ½(-3.5 m/s 2 )t 2 v(t) = (26 m/s) + (-3.5 m/s 2 )t Time bus will come to a stop: 0 m/s = (26 m/s) + (-3.5 m/s 2 )t t = (-26 m/s)/(-3.5 m/s 2 ) = 7.4 s How far bus will go before coming to a stop: x = (26 m/s)(7.4 s) + ½(-3.5 m/s 2 )(7.4 s) 2 = 97 m Assuming constant velocity Reason and Represent a) The object starts 200 m to the left of the origin and is moving in the positive direction at 20 m/s, slowing down at a rate of 2 m/s 2 20 m/s -2 m/s meters 0 meters b) Motion Diagram v 1 v 2 v 3 v 4 c) Position vs. Time Δv 12 Δv 23 Δv 34

15 PUM Physics II - Kinematics Lesson 12 Solutions Page 15 of 16 Velocity vs. Time d) x(t) = (-200 m) + (20 m/s)t + ½(-2 m/s 2 )t 2 v(t) = (20 m/s) + (-2 m/s 2 )t Time when object stops: 0 m/s = (20 m/s) + (-2 m/s 2 )t t = (-20 m/s)/(-2 m/s 2 ) = 10 s Position when object stops: x = (-200 m) + (20 m/s)(10 s) + ½(-2 m/s 2 )(10 s) 2 = -100 m Reason and Represent Answers will vary based on the reference frame chosen a) 6.0 m/s 8.0 s later 9.75 m/s 0 meters b) v 1 v 2 v 3 v 4 v 5 v 6 v 7 Δv 12 Δv 23 Δv 34 Δv 45 Δv 56 Δv 67

16 PUM Physics II - Kinematics Lesson 12 Solutions Page 16 of 16 c) Velocity vs. Time d) Distance Traveled before Acceleration x = (6.0 m/s)(8.0 s) = 48 m Distance Traveled after Acceleration: ( v f! vi ) a = " t a = (9.75 m/s 6.0 m/s)/(5.0 s) = 0.75 m/s 2 x(t) = (6.0 m/s)t + ½(0.75 m/s 2 )t 2 x = (6.0 m/s)(5.0 s) + ½(0.75 m/s 2 )(5.0 s) 2 = 39 m Total Distance: 48 m + 39 m = 87 m Regular Problem a) A car starts at a velocity of 4m/s in the positive direction and accelerates uniformly for 6 seconds until it is at a speed of 22m/s. b) Acceleration of the car: ( v f! vi ) a = " t a = (22 m/s 4 m/s)/(6 s) a = 3 m/s 2 Displacement of the car is equal to the area under the graph line: x(t) = (4 m/s)t + ½(3 m/s 2 )t 2 x = (4 m/s)(6 s) + ½(3 m/s 2 )(6 s) 2 = 78 m c) If the object traveled at a constant velocity of 4 m/s (the velocity at t = 0 s) for 6 s the total displacement would be 24 m. This makes sense since it is less than the displacement for the accelerating object. 24 m is equal to the first term of the x(t) function. It is the displacement if a = 0 m/s 2