Motion Practice Problems with Solutions A video screencast of this presentation can be found at: https://www.youtube.com/watch?v=0XVzo_SJ2js&feature=youtu.be

Motion Graphs Practice Problem Set A The graph on the left depicts the velocities of 4 objects (A,B, C, & D) over a 20 second time interval. Rank the initial velocities (at time 0) for each object from highest to lowest. A (8m/s)>C (4m/s)>D (0m/s)=B (0m/s) Read these values directly off of the Y axis at time 0. Rank the final velocities (at time 20) of each object from highest to lowest. A (8m/s)=B(8m/s)>D (0m/s)=C(0m/s) Read these values directly off of the Y axis at time 20 seconds. Which car ends up farther from its starting point? Justify your answer. To answer this question you need to find the displacement of each object by calculating the area under the curve. A=160 m; B=80 m; C=40 m; D=60 m Since A has the greatest displacement, A is the answer to this question. Which car experiences the greatest acceleration (in terms of magnitude) between t=0 and t=10 seconds. Justify your answer. To answer this question, you need to calculate the slope of each line between t=0 and t=10 seconds. A=0; B=0.4m/s2; C=-0.2m/s2;D=0.6m/s2 D is our answer.

Motion Graphs Practice Problems Set A Continued. What is magnitude of object A’s acceleration between t=0 and t=20? To calculate average acceleration, we must calculate the slope of the line. In this case we get: (8-8)/(20-0)=0 The line is horizontal, therefore the acceleration is 0. What is the magnitude of object C’s acceleration between t=0 and t=20? (0-4)/(20-0)= -0.2m/s2 Which objects move backward during the time interval? None of the objects move backward during the time interval. All of the objects have either a positive or zero value for their velocities during the entire time interval. Describe the change in object D’s velocity that occurred at t=10 seconds. At t=10 seconds, the slope of line D instantaneously changed from a positive value to a negative value. This indicates that the acceleration changed from a positive value to a negative value at that instant. Motion Graphs Practice Problems Set A Continued.

Motion Graphs Practice Problems Set A Continued What is the instantaneous velocity of Car B at t=10 seconds? To determine the instantaneous velocity from a velocity vs. time graph, simply read the value off of the Y axis. See the bright yellow lines include on the graph. The velocity at t= 10 seconds is 4 m/s. What is the total displacement of object D during the 20 second interval? To find the total displacement, calculate the area under the curve. Since the curve forms a triangle, A=1/2 bh=1/2 (20 s) (6m/s) =60 meters

Motion Graphs Practice Problem Set B The graph on the left depicts the position of an object against time from t=0 to t=55 seconds. Rank the speed of the objects in each of the four labeled regions from fastest to slowest. Justify your answer. To determine the velocity of an object from a position vs. time graph, calculate the slope of the line. Since this question only asks for speed, the direction (sign) doesn’t matter. A(6m/s)>C(4m/s)>D(2.67m/s)>B (0m/s) What total distance did the object travel during the 55 second interval? Distance is a scalar and doesn’t factor in direction. The total distance traveled was 60m + 0M + 100M+ 40M=200M What was the total displacement of the object from t=0 to t=55 seconds? Displacement is a vector and does factor in direction. The object ends at the same position it began. Its total displacement is 0. What is the value of the velocity of the object during section C of the graph? To determine the velocity of an object from a position vs. time graph, calculate the slope of the line. Slope=(60- (-40))/(15-40)= -4 m/s In this case the negative sign indicates direction. A D C B

Motion Graphs Practice Problem Set B During which time intervals is the object moving backwards? Justify your response. To determine the velocity of an object from a position vs. time graph, calculate the slope of the line. A(+6m/s)>C(-4m/s)>D(+2.67m/s)>B (0m/s) The only interval in which the object has a negative velocity (is moving backward) is C. What is the displacement of the object from t=15 to t=40? To find the displacement from a position vs. time graph, subtract the initial position from the final position. Δx=x-x0= -40m-60m= -100 meters What is the displacement of the object from t=40 to t=55 seconds? Δx=x-x0= 0m-(-40m)= 40 meters Motion Graphs Practice Problem Set B

Multiple Choice Practice Question 1 The object is moving to the left (negative velocity). As it is moving its speed decreases, therefore it is accelerating to the right (positive).

Multiple Choice Practice Question 2 The object is moving to the right and moves 2 meters during each 1 second time interval. It has a constant velocity of 2m/s and therefore has 0 acceleration.

Multiple Choice Practice Question 3 Once the rock leaves the boy’s hand, it accelerates at the acceleration due to gravity (g)=-9.81m/s2

Multiple Choice Practice Question 4 Graph “a” is a position vs. time graph with a constant slope. Since the slope of this type of graph is equal to the velocity, this indicates that the velocity is constant. Graph “b” is a velocity vs. time graph with zero slope. Since the slope of this type of graph is equal to acceleration, this indicates that the velocity is constant and the acceleration is zero. Graphs “c” and “d” are both velocity vs. time graphs with non zero slopes. This indicates that both objects are accelerating.

Multiple Choice Practice Question 5 This is a velocity vs. time graph. The truck is initially traveling at a faster velocity than the car and continues to do so all the way to time T. Even though the car has a greater acceleration, the truck has a greater velocity and travels a greater distance.

Multiple Choice Practice Question 6 The graph indicates that the particle has a constant, positive acceleration. Since it initially has a negative velocity (-2.5m/s), its speed (direction/sign doesn’t matter since speed is a scalar) slows to 0 at around time 3.5 seconds. Since the acceleration is positive, from t=3.5 to t=9.0 the speed increases from 0 to 4m/s.

Multiple Choice Practice Question 7 Once the ball is thrown, its acceleration is always the acceleration due to gravity (g) and thus always acts in the downward direction.

Multiple Choice Practice Question 8 The easiest way to solve this problem is to use the quadratic equation. First, use the kinematic equation: Δx=v0t + ½ at2 Plug in all of the values from the problem and move everything to one side so that the formula is in the form of a quadratic. This results in: -4.9t2 + 13t – 4=0 Solve for t using the quadratic equation. You get two answers: 0.36 seconds and 2.3 seconds The 0.36 second point indicates the time at which the object passes through the 4.0 meter mark on the way up, while the 2.3 second point indicates the time at which the object passes through the point on the way down.

Multiple Choice Practice Question 9 Use a=(v-v0)/Δt to the solve this problem. This results in: (-26 m/s – 30 m/s)/0.020 seconds=-2800 m/s2

Multiple Choice Practice Question 10 For the Coyote to catch the Road Runner their displacements must be equal. Note: The Coyote must have a faster velocity than the Road Runner since the Coyote started from rest. To solve the problem set the displacements equal. Δxcoyote= ΔxRoad Runner v0t + ½ at2 = vt 0 + ½ a (10s)2 = 55m/s (10s) Solve for “a” to get 11.0m/s2

Multiple Choice Practice Question 11 First, convert 20.0 dam/s to m/s. This yields 200. m/s as the initial velocity. Use the kinematic equation Δx=v0t + ½ at2 to calculate the displacement for the entire 5.00 sec time interval. Δx=200. m/s (5.00 s) + ½ (20.0 m/s2) (5.00s)=1250 m Next calculate the displacement for the first 4.00 seconds. Δx=200. m/s (4.00 s) + ½ (20.0 m/s2) (4.00s)=960. m To calculate the displacement during the fifth second, subtract the displacement for the first 4 seconds from the displacement for all 5 seconds. Δx=1250 m – 960. m=290. meters

Multiple Choice Practice Question 12 When using a motion detector, a positive velocity indicates that the object is moving away from the detector. Since the velocity and the acceleration have opposite signs, the object is slowing down. The correct answer is D.

FRQ Practice Problem 1 A racquetball is thrown horizontally and to the left (negative x direction) against a wall so that it bounces directly back after hitting the wall. (Ignore the small vertical motion of the ball) The horizontal position of the ball’s center of mass (as a function of time) is described in the data table below. 1. Create a position vs. time graph for the data. 2. What was the ball’s velocity (initial) before it made contact with the wall? 3. What was the ball’s velocity (final) after it bounced off the wall? 4. What is the ball’s average acceleration during its contact with the wall (t=6 to 16 milliseconds)? 5. What is the ball’s instantaneous velocity at t=10 milliseconds? Determine this from the graph and describe the process you used to arrive at your answer.

FRQ Practice Problem 1 Answers 1. Graph.

FRQ Practice Problem 1 Answers 2. 3.

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FRQ Practice Problem 2 A dynamics cart (with essentially frictionless wheels) is released from the top of an inclined plane. The cart accelerates down the plane and then continues to roll across a horizontal floor located at the bottom of the plane. Your teacher assigns you the task of developing an experimental procedure that will allow you to measure the car’s acceleration on the ramp and its velocity across the floor. 1. What materials commonly found in a science lab or classroom will you need to conduct this experiment? Explain how you will use each piece of equipment. (You do not have access to computers or computer-based measuring devices such as motion detectors, photogates, smart pulleys, or other probeware.) 2. Describe the procedure, in a step by step manner, that you will use to measure the acceleration of the cart down the plane and the velocity of the cart across the floor. 3. Create an appropriate data table, with clearly labeled headers, that you would use to record your data. 4. Describe how you will use the data to calculate the acceleration of the cart down the plane and the velocity of the cart across the floor.

FRQ Practice Problem 2 Answers 1. 2.

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