Unit 1 · Topic 1.2

1.2 Displacement, Velocity, and Acceleration

Displacement, velocity and acceleration describe where something is, how fast its position changes and how fast its velocity changes. In Physics C you find instantaneous velocity and acceleration as derivatives, $v_x = \frac{dx}{dt}$ and $a_x = \frac{dv_x}{dt}$ , so you can handle motion where the acceleration isn't constant.

Key terms

displacement
average velocity
instantaneous velocity
average acceleration
instantaneous acceleration
derivative

Displacement

Position, x, is where an object is relative to an origin you choose. Displacement is the change in position, $\Delta x = x_f - x_i$ . It depends only on where you start and end, not on the path between. In kinematics you usually treat the moving thing as an object: a single point with mass, ignoring its size, shape and what's going on inside it.

Displacement can be positive, negative or zero. A runner who starts at x = 2 m, runs to x = 10 m and comes back to x = 6 m has a displacement of +4 m, but traveled a distance of 8 + 4 = 12 m.

Average and instantaneous velocity

Average velocity is displacement divided by the time it took:

$v_{avg} = \frac{\Delta x}{\Delta t}$

Average speed is different: it's the total distance divided by the time. If you return to your starting point, your average velocity is zero even though your average speed isn't.

Instantaneous velocity is the velocity at one moment. You get it by shrinking the time interval toward zero, which is exactly the definition of a derivative:

$v_x = \lim_{\Delta t \to 0}\frac{\Delta x}{\Delta t} = \frac{dx}{dt}$

On a position–time graph, average velocity is the slope of the line joining two points (a secant line), and instantaneous velocity is the slope of the tangent line at one point. The speedometer in a car reads instantaneous speed, the magnitude of instantaneous velocity.

Acceleration

Acceleration is the rate at which velocity changes. Average acceleration is $a_{avg} = \frac{\Delta v}{\Delta t}$ , and instantaneous acceleration is the derivative $a_x = \frac{dv_x}{dt} = \frac{d^2x}{dt^2}$ .

Because velocity is a vector, an object accelerates whenever its speed changes or its direction changes. A car going around a curve at a steady 20 m/s is accelerating.

In one dimension, compare signs to tell whether an object is speeding up or slowing down:

Velocity and acceleration have the same sign: the object is speeding up.
Velocity and acceleration have opposite signs: the object is slowing down.
Velocity is zero but acceleration isn't: the object is momentarily at rest and may be turning around.

Working from a position function

Physics C questions often give position as a function of time, such as $x(t) = 2t^3 - 9t^2 + 12t$ . Differentiate once for velocity and again for acceleration. To find when the object is at rest, set $v(t) = 0$ and solve. To find when it changes direction, check that the velocity actually changes sign there.

To find the total distance traveled, find every time the velocity changes sign, compute the position at each of those times, and add up the sizes of each leg. Total distance is never less than the size of the displacement.

Keep track of units: if x is in meters and t in seconds, then v is in m/s and a is in m/s². In $x(t) = 2t^3$ , the 2 secretly has units of m/s³ so the product comes out in meters.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Velocity and acceleration from x(t)
A particle moves along the x-axis with position $x(t) = 2t^3 - 9t^2 + 12t$ , with x in meters and t in seconds. (a) When is the particle at rest? (b) Find its average velocity and average speed from t = 0 to t = 3 s. (c) At t = 1.8 s, is it speeding up or slowing down?
Show the solutionHide the solution
1. Step 1: Differentiate: $v(t) = 6t^2 - 18t + 12 = 6(t-1)(t-2)$ m/s, and $a(t) = 12t - 18$ m/s².
2. Step 2: (a) Set v = 0: the particle is at rest at t = 1 s and t = 2 s. The velocity changes sign at each, so it turns around both times.
3. Step 3: (b) Positions: x(0) = 0, x(1) = 5 m, x(2) = 4 m, x(3) = 9 m. Average velocity = (9 − 0)/3 = 3 m/s.
4. Step 4: Distance: it moves 5 m forward, 1 m back, then 5 m forward, for 11 m total. Average speed = 11/3 ≈ 3.7 m/s, which is bigger than the average velocity because of the backtracking.
5. Step 5: (c) v(1.8) = 6(0.8)(−0.2) = −0.96 m/s and a(1.8) = 12(1.8) − 18 = 3.6 m/s². The signs are opposite, so it's slowing down.
Answer: (a) t = 1 s and t = 2 s; (b) average velocity 3 m/s, average speed about 3.7 m/s; (c) slowing down.
Example 2Calculator allowed
Average acceleration when direction reverses (classic trap)
A car moving east at 20 m/s brakes, stops, and then backs up, reaching 10 m/s west. The whole change takes 4.0 s. Find the car's average acceleration.
Show the solutionHide the solution
1. Step 1: Take east as positive. Then $v_i = +20$ m/s and $v_f = -10$ m/s.
2. Step 2: $\Delta v = v_f - v_i = -10 - 20 = -30$ m/s. The trap is computing 10 − 20 = −10 m/s by ignoring the reversal.
3. Step 3: $a_{avg} = \frac{\Delta v}{\Delta t} = \frac{-30}{4.0} = -7.5$ m/s².
4. Step 4: The negative sign means the average acceleration points west the whole time, even though the car first slows down and then speeds up.
Answer: 7.5 m/s² west (−7.5 m/s² with east positive).

Common mistakes

Mixing up average speed and average velocity. Average velocity uses displacement; average speed uses total distance.
Saying a negative acceleration always means slowing down. It only means slowing down when the velocity is positive.
Assuming an object changes direction everywhere v = 0. Check that v actually changes sign, not just touches zero.
Using constant-acceleration equations on a position function like $x = 2t^3 - 9t^2 + 12t$ . Its acceleration changes, so differentiate instead.

On the exam

Expect a position, velocity or acceleration function and questions about when the object is at rest, turns around, or speeds up. Show the derivative you took; it earns points on free response.
Questions often contrast an average value over an interval with an instantaneous value at a moment. Read carefully for which one is asked.

Connected topics

Videos

Kinematic quantities: Rates of change | AP Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Topic 1.2 - Displacement, Velocity and Acceleration
Lessons With LondotWatch on YouTube (opens in a new tab)
The Derivative and a Demonstration of Position, Velocity and Acceleration
Flipping PhysicsWatch on YouTube (opens in a new tab)
AP Physics C - Defining Motion
Dan Fullerton (APlusPhysics)Watch on YouTube (opens in a new tab)
AP Physics C Mechanics - Unit 1 - Lesson 6C - Calculus with Kinematics
Allen Tsao The STEM CoachWatch on YouTube (opens in a new tab)
1.7 Worked Example: Derivatives in Kinematics
MIT OpenCourseWareWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 1.2 Displacement, Velocity, and Acceleration. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A particle's position along the x-axis is $x(t) = 2t^3 - 6t^2 + 3$ , with x in meters and t in seconds. What is its velocity at t = 3 s?

Question 2 of 4Calculator allowed

A cart's acceleration is a(t) = 6t, with a in m/s² and t in seconds. The cart's velocity at t = 0 is 2 m/s. What is its velocity at t = 2 s?

Question 3 of 4Calculator allowed

A car is moving at 20 m/s in the +x direction. Five seconds later it is moving at 10 m/s in the −x direction. What is its average acceleration over those 5 s?

Question 4 of 4Calculator allowed

In which situation is the object's acceleration zero?

0 of 4 answered