AP Physics 1 · Lesson 2

Motion graphs and constant acceleration

Know Know it

On a position-time graph, the slope is the velocity: a straight line means constant velocity and a curve means the velocity is changing. On a velocity-time graph, the slope is the acceleration and the area between the line and the time axis is the displacement, with area below the axis counting as negative.

When the acceleration is constant, three equations link the five quantities v0v_0, vv, aa, tt and Δx\Delta x: v=v0+atv = v_0 + at; Δx=v0t+12at2\Delta x = v_0t + \tfrac{1}{2}at^2; v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x. Each one leaves out one quantity, so pick the equation that leaves out the quantity you neither know nor need.

Free fall is constant acceleration too. With up as positive, a=−9.8a = -9.8 m/s² on the way up, at the top and on the way down.

What each motion graph tells you
GraphSlope givesArea under the line gives
Position vs. timeVelocityNothing useful
Velocity vs. timeAccelerationDisplacement
Acceleration vs. timeNot needed in this courseChange in velocity

Worked example Easy

A city bus pulls away from a stop, starting from rest with a constant acceleration of 1.5 m/s². What is its speed after 8.0 s?

  1. 5.3 m/s
  2. 6.0 m/s
  3. 12 m/sAnswer
  4. 48 m/s

How to solve it

  1. Knowns: v0=0v_0 = 0, a=1.5a = 1.5 m/s², t=8.0t = 8.0 s. Wanted: v.
  2. v=v0+atv = v_0 + at is the equation without Δx\Delta x.
  3. v=0+(1.5)(8.0)=12v = 0 + (1.5)(8.0) = 12 m/s.

Why each choice is right or wrong

  • A. Incorrect. 5.3 m/s divides the time by the acceleration (8.0/1.5); the speed gained is the acceleration times the time.
  • B. Incorrect. 6.0 m/s is the bus’s average speed over the 8.0 s, half its final speed, not its speed at 8.0 s.
  • C. Correct. v=v0+at=0+(1.5)(8.0)=12v = v_0 + at = 0 + (1.5)(8.0) = 12 m/s.
  • D. Incorrect. 48 m/s is the distance in meters the bus covers, 12(1.5)(8.0)2=48\tfrac{1}{2}(1.5)(8.0)^2 = 48 m, not its speed.

Trap Zero velocity is not zero acceleration

At the top of a straight-up toss, or at the instant a cart reverses, the velocity is zero for an instant but the acceleration is not. On a velocity-time graph this is a line crossing the time axis: its value there is zero, but its slope is not.

Explain Explain it

Task verbs: calculate, interpret

To interpret a graph, say what its slope and area mean in words, with units. A slope of -2 m/s² on a velocity-time graph means the object loses 2 m/s of velocity every second. Where the line crosses the time axis, the object stops for an instant and turns around.

In equation problems, list the knowns with their signs first. An object that is slowing down has an acceleration opposite to its velocity, so if the velocity is positive the acceleration is negative.

Worked example Medium

A scooter moving at 9.0 m/s slows down at a constant rate to 3.0 m/s while covering 24 m. What is its acceleration?

  1. -3.0 m/s²
  2. -1.5 m/s²Answer
  3. -0.25 m/s²
  4. +1.5 m/s²

How to solve it

  1. Knowns: v0=9.0v_0 = 9.0 m/s, v=3.0v = 3.0 m/s, Δx=24\Delta x = 24 m. The time is neither given nor needed.
  2. Use v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x, the equation without t.
  3. a=(3.02−9.02)/(2×24)=−72/48=−1.5a = (3.0^2 - 9.0^2)/(2 \times 24) = -72/48 = -1.5 m/s².

Why each choice is right or wrong

  • A. Incorrect. -3.0 m/s² leaves out the 2 in v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x, so it is twice the true size.
  • B. Correct. a=(v2−v02)/(2Δx)=(9.0−81)/48=−1.5a = (v^2 - v_0^2)/(2\Delta x) = (9.0 - 81)/48 = -1.5 m/s², negative because the scooter slows while moving in the positive direction.
  • C. Incorrect. -0.25 m/s² divides the change in speed by the distance; acceleration is the change in velocity per second, not per meter.
  • D. Incorrect. +1.5 m/s² has the right size but the wrong sign: the scooter is slowing down, so its acceleration points against its motion.

Interpret Your turn

An elevator’s velocity-time graph (up is positive) has three straight segments: the velocity rises from 0 to 2.0 m/s between t = 0 and t = 2.0 s, stays at 2.0 m/s until t = 8.0 s, and then falls back to 0 at t = 10.0 s. Find the elevator’s displacement over the 10 s, and find its acceleration during the first 2.0 s and during the last 2.0 s.

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Compare with a model answer

A full answer does this

  • Gives the displacement over the 10 s as 16 m upward
  • Gives the acceleration during the first 2.0 s as +1.0 m/s²
  • Gives the acceleration during the last 2.0 s as -1.0 m/s²

Model answer

The displacement is the area under the graph: a triangle of 12(2.0)(2.0)=2.0\tfrac{1}{2}(2.0)(2.0) = 2.0 m, a rectangle of (6.0)(2.0) = 12 m and another triangle of 2.0 m, so 16 m upward. In the first 2.0 s the slope is 2.0 - 0/2.0 = +1.0 m/s², so the acceleration is 1.0 m/s² upward. In the last 2.0 s the slope is 0 - 2.0/2.0 = -1.0 m/s², so the acceleration is 1.0 m/s² downward, even though the elevator is still moving up.

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