# AP Physics: Displacement, Velocity and Acceleration

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 Answer

**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 Answer

**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.

### 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 Answer

**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 origin, any portion of the graph moving away from the origin is moving away from the home, moving toward the origin is toward the home, and any portion with zero slope represents no motion.

Question 2 Answer

**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**

Question 3 Answer

**12 blocks; see graph below**

- 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 4blocks forward, for a total distance of 12 blocks.
- The key difference in a distance vs. time graph is that distance will only move upwards over time:

Question 4 Answer

**2 blocks/min, 0 blocks/min, –3 blocks/min, 0 blocks/min, 2 blocks/min****0 blocks/min****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 Answer

**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 Answer

**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 Answer

**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.

Question 3 Answer

**See graphs below.**

1. The ball will start at 20 m, accelerate towards the ground, then follow the exact reverse path up to a height of 20 m again after bouncing.

2. The ball’s velocity starts at zero, increases linearly in a downward direction until it bounces, at which point it suddenly (and discontinuously) jumps to a high value in the upward direction, then slows to zero again.

3. The ball’s acceleration holds constant at g (9.8 m/s^2 downwards) until it bounces. In the split-second that it bounces, acceleration spikes to a very high value; then it immediately returns to g once again.