Practice Problems: Motion Graphs
1. (easy) The graph below shows how a car moves along the straight road as a function of time. Describe (in words) what the car is doing during the experiment and what devices on the car must be utilized for the motion to occur.
At first the car is at rest, some distance behind the zero point in the frame of reference (FOR). The car then begins to move forward at a constant speed. Since the transition from rest to constant speed appears to be instantaneous, we assume a very quick acceleration with the gas pedal and then the driver throttles back on the gas pedal but still uses it to maintain a steady speed. She could also use the cruise control on the car to maintain the speed. At some distance in front of the zero point of the FOR the driver uses the brakes to stop the car very quickly (instantaneously in the graph). The car remains at rest at that position for a period of time. The driver then puts the car in reverse and instantaneously achieves a constant negative velocity (again using the gas pedal) for a shorter period of time than any other part of the motion. The car is still moving in reverse when it reaches the origin of the FOR.
2. (easy) If the time axis in question #1 was shifted upward how would that change your response to the question?
The shift of the time axis to a higher position would not change the description of the motion for each segment, but it would change the position of the object relative to the origin of the FOR at any particular moment.
3. (moderate) By analyzing the x-t graph shown here, answer the questions below.
The first segment of the graph has a width of 5 seconds and a height of 7 m. The width of the middle segment of the graph is 6 seconds. The final segment of the graph has a width of 3 seconds.
A. Create an accurate v-t graph that expresses the same motion.
B. What is the velocity of the object during the first 5 seconds?
v = Δx/Δt = 7m/5s = 1.4 m/s
C. What is the average velocity of the object over the entire experiment?
v = Δx/Δt = 0m/14s = 0 m/s
D. What is the average speed of the object over the entire experiement?
v = Δx/Δt = 14m/14s = 1.0 m/s
E. What is the average velocity during the final three seconds?
v = Δx/Δt = -7m/3s = -2.3 m/s
F. Displacement can be determined on a v-t graph by studying what are called the areas "under the curve". In the case of a linear graph this still refers to the areas above and below the lines. The area always represents a velocity multiplied by a time. Hence, the area is a displacement. If the area is above the time axis, the displacement is positive. If it is below the time-axis, the displacement is negative.What is the total area under the v-t graph you created? How does that answer correlate to the x-t graph?
The total area is zero. The area in the first segment (7 m) cancels with the area in the final segment (-7 m). This corresponds to the x-t graph because the object starts and ends its motion at x = 0.
4. (moderate) By analyzing the v-t graph shown here, answer the questions below.
The initial velocity on the graph is 10 m/s forward. It takes 8 seconds for the object to come to rest (momentarily.) The object moves backward for an additional 2 seconds until it reaches a final velocity of 2.5 m/s backward.
A. How far does the object move in the first 8 seconds?
Displacement is the area under the curve. In this case, it represents the area of the triangle that lies above the time axis.
Area = displacement = ½(base)(height) = ½(8s)(10m/s) = 40m
B. If the initial position of the object was at x = -15 meters, what is the final position?The overall displacement of the object is the area under both triangles. As shown in the previous work, the displacement for the first 8 seconds is +40m. Likewise, one can calculate the area under the small triangle to be -2.5m. Thus the overall displacement is +37.5m.
Δx = 37.5m = xf - xo = xf - (-15 m) Therefore, xf = 22.5 m
C. What is the acceleration of this object?
The acceleration is the slope of the v-t graph. Since the slope is constant in this case, the instantaneous acceleration is the same as the average acceleration over the entire experiment.
a = Δv/Δt = [-2.5m/s - 10m/s]/10s = -1.25m/s2
5. (moderate) If the time axis on the v-t graph in the previous question was shifted slightly upward, how would that change the magnitude of your answers?
A. The area inside the initial segment of the motion would be smaller, because it would take less time for the obect to come to rest. Additionally, there would be some backward motion prior to t = 8 seconds. Thus the displacement for the first 8 seconds would have a smaller positive magnitude.
B. The area of the second segment of the motion would increase (more negative displacement) and cause the overall final position to be slightly less positive.
C. Since the position of the time axis does not affect the slope of the line on the graph, and the slope is the acceleration, the answer in would not change.
6. (moderate) Consider the motion of the horse in the sketch below. During the first 10 seconds the horse moved with constant speed. During the next 10 seconds the horse accelerated uniformly. During the final 20 seconds the horse underwent a different uniform acceleration. Create a motion graph (position vs. time) for the horse.
7. (moderate) Solve the problem shown below.