Unit 1 · Topic 1.3

1.3 Representing Motion

The same motion can be described with a graph, an equation, a diagram or words, and the exam asks you to move between them. Slopes and areas connect the position, velocity and acceleration graphs, integrals undo derivatives, and the kinematic equations handle the special case of constant acceleration, like free fall.

Key terms

position–time graph
velocity–time graph
slope
area under the curve
kinematic equations
free fall

Slopes and areas

Three graphs describe one-dimensional motion, and they're linked in two directions. Going down the chain (position → velocity → acceleration) you take slopes, which are derivatives. Going up the chain you take areas, which are integrals.

A motion diagram is another representation: dots showing the object's position at equal time intervals, like a strobe photo. Evenly spaced dots mean constant velocity, spreading dots mean speeding up, and bunching dots mean slowing down.

Graph	Slope gives	Area under it gives
position vs. time	velocity	(not used)
velocity vs. time	acceleration	displacement
acceleration vs. time	(not used)	change in velocity

Integrals take you back up

If you know velocity as a function of time, the displacement between two times is the area under the velocity curve:

$\Delta x = \int_{t_1}^{t_2} v_x\,dt$

Similarly, $\Delta v_x = \int_{t_1}^{t_2} a_x\,dt$ . Area below the time axis counts as negative. To get a position (not just a change), add the starting value: $x(t) = x_0 + \int_0^t v_x\,dt$ . Forgetting the initial position or initial velocity is the same as forgetting the constant of integration.

Reading graphs: on a position graph, a curve that bends upward (concave up) means positive acceleration. On a velocity graph, the object stops where the curve crosses the time axis. A horizontal velocity graph means constant velocity, and the position graph is then a straight sloped line.

The constant-acceleration equations

When the acceleration is constant, integrating twice gives the kinematic equations. Each one leaves out one variable, so pick the one that skips the quantity you don't know and don't need:

$v_x = v_{x0} + a_xt$ (no displacement)
$x = x_0 + v_{x0}t + \frac{1}{2}a_xt^2$ (no final velocity)
$v_x^2 = v_{x0}^2 + 2a_x(x - x_0)$ (no time)

Free fall

An object in free fall moves under gravity alone, so near Earth's surface its acceleration is g ≈ 9.8 m/s² downward, whatever its mass and whether it's moving up, down or sitting at the top of its path. The course description says exam questions that need a number for g use 10 m/s², and you won't lose credit for using 9.8 or 9.81 m/s² (the equation sheet lists 9.8). If a problem gives a value, use that one.

For a ball thrown straight up (up positive), the position graph is a downward-opening parabola, the velocity graph is a straight line with slope −g that crosses zero at the top, and the acceleration graph is a flat line at −g.

When the equations don't apply

If the acceleration changes with time, like $a = 6t$ , the kinematic equations give wrong answers. Integrate instead. A quick test: if the problem gives a(t), v(t) or a force that depends on time or speed, reach for calculus.

Motion in pieces is fine, though. A car that speeds up steadily and then cruises has constant acceleration in each stage. Use the kinematic equations stage by stage, with the final velocity of one stage as the starting velocity of the next.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Integrating a changing acceleration
A cart's acceleration is $a(t) = 6t$ m/s². At t = 0 it is at x = 1.0 m moving at 2.0 m/s. Find its velocity and position at t = 2.0 s.
Show the solutionHide the solution
1. Step 1: The acceleration changes with time, so the kinematic equations don't apply. Integrate.
2. Step 2: $v(t) = v_0 + \int_0^t 6t\,dt = 2 + 3t^2$ . At t = 2.0 s, $v = 2 + 3(4) = 14$ m/s.
3. Step 3: $x(t) = x_0 + \int_0^t (2 + 3t^2)\,dt = 1 + 2t + t^3$ . At t = 2.0 s, $x = 1 + 4 + 8 = 13$ m.
4. Step 4: A check you can't do with $\frac{1}{2}at^2$ : using the t = 2 acceleration of 12 m/s² as if it were constant would give 1 + 4 + 24 = 29 m, which is far too big.
Answer: v = 14 m/s and x = 13 m at t = 2.0 s.
Example 2Calculator allowed
A ball thrown upward off a ledge
A ball is thrown straight up at 15 m/s from a point 2.0 m above the ground. Ignore air resistance and use g = 9.8 m/s². Find (a) its maximum height above the ground, (b) the time until it hits the ground and (c) its speed just before impact.
Show the solutionHide the solution
1. Step 1: Take up as positive, so a = −9.8 m/s², with y₀ = 2.0 m and v₀ = 15 m/s.
2. Step 2: (a) At the top v = 0. Use the equation without time: $0 = 15^2 - 2(9.8)(y - 2.0)$ , so $y - 2.0 = \frac{225}{19.6} = 11.5$ m and the maximum height is about 13.5 m.
3. Step 3: (b) Set y = 0 in $y = y_0 + v_0t - \frac{1}{2}gt^2$ : $0 = 2.0 + 15t - 4.9t^2$ . The quadratic formula gives t ≈ 3.19 s (the negative root isn't physical).
4. Step 4: (c) $v^2 = 15^2 + 2(9.8)(2.0) = 264.2$ , so the speed is about 16.3 m/s. It's a bit faster than 15 m/s because it falls 2.0 m below its launch point.
5. Step 5: The acceleration at the top is still 9.8 m/s² downward. Only the velocity is zero there.
Answer: (a) about 13.5 m; (b) about 3.19 s; (c) about 16.3 m/s.
Example 3
Displacement from a velocity graph
A velocity–time graph rises in a straight line from 0 to 8.0 m/s between t = 0 and t = 4.0 s, stays at 8.0 m/s until t = 6.0 s, then falls in a straight line to 0 at t = 10.0 s. Find the total displacement and the acceleration during each part.
Show the solutionHide the solution
1. Step 1: Displacement is the area under the graph. Split it into shapes.
2. Step 2: 0–4 s: triangle, $\frac{1}{2}(4.0)(8.0) = 16$ m. 4–6 s: rectangle, (2.0)(8.0) = 16 m. 6–10 s: triangle, $\frac{1}{2}(4.0)(8.0) = 16$ m. Total = 48 m.
3. Step 3: Accelerations are the slopes: 8.0/4.0 = 2.0 m/s², then 0, then −8.0/4.0 = −2.0 m/s².
Answer: 48 m; accelerations of 2.0 m/s², 0 and −2.0 m/s².

Common mistakes

Using the kinematic equations when acceleration isn't constant. Check first; if a depends on t or v, integrate.
Forgetting the starting position or starting velocity after integrating. They play the role of the constant of integration.
Reading the height of a velocity graph as position. On a v–t graph, position is the area, not the height.
Saying the acceleration is zero at the top of a toss. Only the velocity is zero; the acceleration is still g downward.

On the exam

Translation Between Representations questions often have you sketch one graph from another. Get the key features right: where slopes are zero, where curves cross the axis, and which parts are straight or curved.
When you integrate on a free-response question, write the integral with its limits before plugging in numbers. That setup often earns a point by itself.

Connected topics

Videos

Kinematic quantities: Accumulations of change | AP Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Topic 1.3 - Representing Motion
Lessons With LondotWatch on YouTube (opens in a new tab)
AP Physics C: Integrals in Kinematics Review (Mechanics)
Flipping PhysicsWatch on YouTube (opens in a new tab)
Velocity and Position From Acceleration By Integration - Physics and Calculus
The Organic Chemistry TutorWatch on YouTube (opens in a new tab)
Developing kinematic equations from data | AP Physics | Khan Academy
Khan AcademyWatch on YouTube (opens in a new tab)
Motion Graphs - AP Physics 1: Unit 1 Review Supplement
Flipping PhysicsWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 1.3 Representing Motion. Pick an answer to see if you got it, and why.

Time (s)	Velocity (m/s)
0	0
1	2.0
2	4.0
3	6.0
4	6.0
5	6.0
6	3.0

Experimental data: a cart moves along a straight track. Its velocity changes at a steady rate between each pair of listed times, so a graph of velocity versus time is made of straight segments joining the points.

Question 1 of 4Calculator allowed

What is the cart's acceleration between t = 0 and t = 3 s?

Question 2 of 4Calculator allowed

How far does the cart travel from t = 0 to t = 6 s?

Question 3 of 4Calculator allowed

What is the cart's acceleration at t = 5.5 s?

Question 4 of 4Calculator allowed

A car moving at 25 m/s brakes with a constant acceleration of magnitude 5.0 m/s² until it stops. How far does it travel while braking?

0 of 4 answered