AP Physics: Displacement, Velocity and Acceleration

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Displacement, velocity, and acceleration are three fundamental physics topics. Mastering these will give you a big head start on some of the later, more complex topics! Take a look at these objectives, notes, and practice questions to check your understanding.


Objective 1


Describe the relationship between displacement, velocity, and acceleration

Objective 1 Notes

  • Distance and displacement are related, but different
    • Both distance and displacement indicate how far an object has moved
    • Because distance is a scalar, distance increases whenever an object moves in any direction
    • Displacement is a vector: Direction matters!
  • A moving object can gain or lose displacement depending on its direction, but a moving object always gains distance
  • Displacement’s advantage is that it shows distance between initial and final position, irrespective of the path taken
  • Speed and velocity are related, but different
    • Both terms indicate the rate at which an object is moving
    • Recall that a “rate” is any number divided by time
    • Speed is distance divided by time; velocity is displacement divided by time
  • Because distance is a scalar, speed (distance/time) is also a scalar
    • Direction does not matter for speed
    • Unlike distance, which always increases or stays the same, speed can increase or decrease over time
    • Like distance, speed is never negative
  • Because displacement is a vector, velocity is also a vector
  • Acceleration is any change in velocity
    • An increase or decrease in velocity’s magnitude is acceleration
    • Also, a change in velocity’s direction is acceleration!

Objective 1 Practice Questions


Question 1
Two friends intend to ascend a mountain. One friend hikes to the top, taking switchbacks back and forth across the mountain. The other flies to the top in a hot air balloon. Which friend has the greater distance traveled? The greater displacement?


The first friend travels the greater distance, but they both have the same displacement. Displacement is a vector quantity, and only depends on the starting and ending points of the object. Distance is a scalar quantity, and so a longer route taken will cause the object to have a greater distance.


Question 2
A car drives east at 40 m/s for 500 seconds, stops, turns around, then drives west at 50 m/s for 400 seconds and stops. What is the displacement of the car? What is the distance the car traveled?


Distance = 4000 m, Displacement = 0 m. In the first leg of the trip, the car travels a distance of 2000 m (40 m/s x 500 s). Then, the car travels the same 2000 m (50 m/s x 400 s) in the opposite direction and returns back to its starting point. Therefore, the total distance covered is 4000 m, while the displacement is zero since the car ended in the same place it started.


Question 3
A car sits idling for five seconds, then accelerates at 4 m/s^2 for five seconds, then continues at that velocity for five seconds. What is the displacement of the car?


150 m. To find the total displacement, add the distances traveled while the car is accelerating and after it reaches its top speed:
AP Physics Practice Question

Objective 2


Gather information from displacement vs. time graphs

Objective 2 Notes

  • Imagine an object moving along a straight line path
  • A displacement versus time graph shows:
    • Either distance or displacement on the y-axis
    • Time on the x-axis
  • The graph’s slope is crucial!
    • If it slopes away from the x-axis, the object moves away from its starting place
    • If it slopes toward the x-axis, the object is moving closer to where it started
    • If it is flat, then the object is stationary
  • Since displacement is related to distance:
    • Distance traveled can be calculated from a displacement vs. time graph
    • Distance vs. time can be plotted from a displacement vs. time graph
  • Slope is always Δy/Δx
    • For displacement vs. time, this is Δdisplacement/Δtime = velocity!
    • You can calculate an object’s velocity by finding the slope of its displacement vs. time graph
    • Where the curve is flat, velocity is zero
  • A displacement vs. time graph also gives information about acceleration
    • If a segment of the curve is linear, then acceleration is zero
    • If a segment of the curve is nonlinear, calculus is required to find acceleration

Objective 2 Practice Questions


Question 1
The graph below shows a woman walking along the street she lives on (home is at zero displacement). Where is the woman walking away from her home? Where is the woman walking toward her home? Where is she stationary?
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Away from home: 0.5 to 1.5 s, 3 to 4 s
Stationary: 1.5 to 2 s, 4 to 4.5 s
Toward home: 2 to 3 s, 4.5 to 5.5 s
Since the woman’s home is at the x-axis, any portion of the graph moving away from the x-axis is moving away from the home, moving toward the x-axis is toward the home, and any portion with zero slope represents no motion.


Question 2
Assume the woman in the graph below walks east and west along the city block she lives on, so the y-axis is in units of city blocks and the x-axis is in units of minutes. Indicate her position at the following time-points: 0, 1.5, 3, 4.5, and 6 minutes.
AP Physics Practice Questions


0 min: 0 blocks
1.5 min: 2 blocks east
3 min: 2.5 blocks west
4.5 min: 3 blocks west
6 min: 0 blocks
Any positive displacement is east while negative displacements are west, with the
displacement represented by the y-value at the given x-value.

Question 3

Using the displacement vs. time graph in question 2:

  1. What is the woman’s total distance traveled?
  2. Convert the graph from a displacement vs. time graph into a distance vs. time graph. (Draw your answer.)


12 blocks; see graph below

  1. The total distance is found by adding the amount of blocks walked in each portion of the graph. The woman walks 2 blocks out, and then 6 blocks backward, and finally 4 blocks forward, for a total distance of 12 blocks.
  2. The key difference in a distance vs. time graph is that distance will only move upwards over time:

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Question 4
Using the displacement vs. time graph in question 2, find:

  1. The woman’s velocity at each point in the curve.
  2. The woman’s total average velocity for the trip.
  3. The woman’s total average speed for the trip.


  1. 2 blocks/min, 0 blocks/min, –3 blocks/min, 0 blocks/min, 2 blocks/min
  2. 0 blocks/min
  3. 2 blocks/min

The velocity at each point in the curve is equal to the slope at that point. The total average velocity for the trip is equal to the displacement over the total time. Since the displacement for the trip is zero, the average velocity is zero as well. Finally, average speed is distance over time, which for this trip is 12 blocks/6 minutes or 2 blocks/min.


Question 5
In the displacement vs. time graph in question 2, what is the woman’s acceleration at the midpoint of each segment of the curve?


0 m/s^2
Since the velocity at each segment of the curve is constant (segments are linear), the acceleration must be zero at each point as well.

Objective 3


Interpret velocity vs. time and acceleration vs. time graphs

Objective 3 Notes

  • A velocity vs. time graph gives information about velocity, displacement, and acceleration!
    • Velocity at a given time can be read directly from the y-axis
    • Acceleration is the slope of a velocity vs. time graph
    • You cannot tell an object’s displacement from a velocity vs. time graph alone, but you can tell its change in displacement
  • To use a velocity vs. time graph to calculate either change in displacement or total distance traveled:
    • Find the area between the curve and the x-axis
    • For displacement, treat area under negative velocities as “negative area”
    • For distance, treat all areas as positive
  • An acceleration vs. time graph gives limited information
    • The graph gives direct information about acceleration
    • You cannot tell velocity directly, though you can find the change in velocity
    • To find the change in velocity, find the area between the curve and the x-axis
    • You cannot tell displacement from an acceleration vs. time graph, though you can make predictions about how displacement is changing

Objective 3 Practice Questions


Question 1
A woman leaves her house and begins walking north. The plot below shows her velocity during her walk.

  1. Indicate her velocity and acceleration at each of the following time-points: 0.5 min, 1 min, 1.5 min, 2 min, 2.5 min.
  2. At what time does she walk south past her house?

AP Physics Practice Questions


  1.  AP Physics Practice Questions
  2. The woman walks past her house at time = 4 minutes.

The velocity is found by reading the value off the graph, and is undefined if there is a vertical line at that point. The acceleration is given by the slope of the graph, and is undefined if the curve changes slope at that point. Finally, the total distance is given by the area under the curve, and the woman passes her home at the point where the total distance is equal to zero again, which is after 4 minutes (areas above the axis are positive, areas below are negative, and at 4 minutes positive and negative areas each equal 1.5 blocks).


Question 2
In the graph below, what is the woman’s net displacement? What is the woman’s total distance traveled?
AP Physics Practice Questions


Distance = 5.5 blocks, displacement = −2 blocks
The distance in each segment can be found as the area under the curve. The total distance is found by adding the areas, and will thus be 1.75 blocks + 3.75 blocks = 5.5 blocks. The displacement, however, takes into account the direction and thus the second area must be subtracted: 1.75 blocks − 3.75 blocks = −2 blocks. If you have trouble calculating the areas of irregular shapes like the ones in this question, try dividing the shaded areas into rectangles and triangles, then calculating the area of each individual rectangle and triangle.

Want more AP Physics practice? Check out more practice questions here!

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