Motion Graphs: Reading Derivatives and Areas

You’ll be able to

Move between position, velocity and acceleration graphs using slopes and areas
Extract displacement from a velocity graph by computing signed area
Distinguish displacement from distance traveled on a graph that changes sign

One relationship, read two directions

Every motion graph question is the same calculus asked in one of two directions. Going down the chain — position to velocity to acceleration — you take slopes, because v = dx/dt and a = dv/dt. Going up the chain you take areas, because Δx = ∫v dt and Δv = ∫a dt. Nothing else is involved. A graph problem that looks unfamiliar is almost always asking you to identify which direction you are moving and then read the appropriate feature off the curve.

Slopes down, areas up

v = dx/dt a = dv/dt Δv = ∫a dt Δx = ∫v dt

A slope has units of the vertical axis divided by the horizontal; an area has units of their product. Checking units is the fastest way to confirm you read the right feature.

Signed area, and the difference between displacement and distance

The integral of velocity gives displacement, and it is a signed area: time spent with v < 0 contributes negatively. Distance traveled is the integral of |v|, so you compute each region separately and add the magnitudes. A particle that moves +24 m and then −24 m has a displacement of zero and a distance of 48 m. Exam questions exploit this difference constantly, and the giveaway is the word used in the prompt.

What curvature tells you

On a position graph, the slope is velocity and the concavity is acceleration: concave up means a > 0. On a velocity graph, the slope is acceleration directly. A common trap is a velocity graph that is positive and decreasing — the object is still moving forward, but slowing, because v > 0 while a < 0. Speeding up or slowing down is decided by whether v and a share a sign, never by the sign of a alone.

Worked example

A particle's velocity increases linearly from 0 to 12 m/s over the interval t = 0 to 2 s, then decreases linearly back to 0 at t = 4 s. Find the acceleration on each interval, the displacement over the full 4 s, and the distance traveled.

1.On 0 to 2 s the slope is a = (12 − 0)/(2 − 0) = 6 m/s².
2.On 2 to 4 s the slope is a = (0 − 12)/(4 − 2) = −6 m/s².
3.Displacement is the area under v(t), a triangle of base 4 s and height 12 m/s: Δx = ½(4)(12) = 24 m.
4.The velocity never becomes negative, so no area is subtracted and the distance traveled equals the displacement, 24 m.

Answer: a = +6 m/s² then −6 m/s²; Δx = 24 m; distance = 24 m, equal to the displacement because v never changes sign

Checkpoint

A velocity–time graph shows v positive and decreasing toward zero. The object is —

On the exam

Read the question for displacement versus distance before computing anything. If any part of the velocity graph lies below the axis the two answers differ, and the difference is usually the entire point of the item.

Checkpoint

On a position–time graph, the acceleration of the object corresponds to —

Answer the 2 checkpoints as you read.