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## AP®︎/College Physics 1

### Course: AP®︎/College Physics 1>Unit 1

Lesson 5: Average and instantaneous acceleration

# What are velocity vs. time graphs?

How to analyze graphs that relate velocity and time to acceleration and displacement.

## What does the vertical axis represent on a velocity graph?

The vertical axis represents the velocity of the object. This probably sounds obvious, but be forewarned—velocity graphs are notoriously difficult to interpret. People get so used to finding velocity by determining the slope—as would be done with a position graph—they forget that for velocity graphs the value of the vertical axis is giving the velocity.
Try sliding the dot horizontally on the example graph below to choose different times and see how the velocity changes.

Concept check: What is the velocity of the object at time t, equals, 4, start text, space, s, e, c, o, n, d, s, end text, according to the graph above?

## What does the slope represent on a velocity graph?

The slope of a velocity graph represents the acceleration of the object. So, the value of the slope at a particular time represents the acceleration of the object at that instant.
The slope of a velocity graph will be given by the following formula:
start text, s, l, o, p, e, end text, equals, start fraction, start text, r, i, s, e, end text, divided by, start text, r, u, n, end text, end fraction, equals, start fraction, v, start subscript, 2, end subscript, minus, v, start subscript, 1, end subscript, divided by, t, start subscript, 2, end subscript, minus, t, start subscript, 1, end subscript, end fraction, equals, start fraction, delta, v, divided by, delta, t, end fraction
Since start fraction, delta, v, divided by, delta, t, end fraction is the definition of acceleration, the slope of a velocity graph must equal the acceleration of the object.
start text, s, l, o, p, e, end text, equals, start text, a, c, c, e, l, e, r, a, t, i, o, n, end text
This means that when the slope is steep, the object will be changing velocity rapidly. When the slope is shallow, the object will not be changing its velocity as rapidly. This also means that if the slope is negative—directed downwards—the acceleration will be negative, and if the slope is positive—directed upwards—the acceleration will be positive.
Try sliding the dot horizontally on the example velocity graph below to see what the slope looks like for particular moments in time.

The slope of the curve is positive between the times t, equals, 0, start text, space, s, end text and t, equals, 2, start text, space, s, end text since the slope is directed upward. This means the acceleration is positive.
The slope of the curve is negative between t, equals, 2, start text, space, s, end text and t, equals, 8, start text, space, s, end text since the slope is directed downward. This means the acceleration is negative.
At t, equals, 2, start text, space, s, end text, the slope is zero since the tangent line is horizontal. This means the acceleration is zero at that moment.
Concept check: Is the object whose motion is described by the graph above speeding up or slowing down at time t, equals, 4, start text, space, s, end text?

## What does the area under a velocity graph represent?

The area under a velocity graph represents the displacement of the object. To see why, consider the following graph of motion that shows an object maintaining a constant velocity of 6 meters per second for a time of 5 seconds.
To find the displacement during this time interval, we could use this formula
delta, x, equals, v, delta, t, equals, left parenthesis, 6, start text, space, m, slash, s, end text, right parenthesis, left parenthesis, 5, start text, space, s, end text, right parenthesis, equals, 30, start text, space, m, end text
which gives a displacement of 30, start text, space, m, end text.
Now we're going to show that this was equivalent to finding the area under the curve. Consider the rectangle of area made by the graph as seen below.
The area of this rectangle can be found by multiplying height of the rectangle, 6 m/s, times its width, 5 s, which would give
start text, space, a, r, e, a, end text, equals, start text, h, e, i, g, h, t, end text, times, start text, w, i, d, t, h, end text, equals, 6, start text, space, m, slash, s, end text, times, 5, start text, space, s, end text, equals, 30, start text, space, m, end text
This is the same answer we got before for the displacement. The area under a velocity curve, regardless of the shape, will equal the displacement during that time interval.
start text, a, r, e, a, space, u, n, d, e, r, space, c, u, r, v, e, end text, equals, start text, d, i, s, p, l, a, c, e, m, e, n, t, end text

## What do solved examples involving velocity vs. time graphs look like?

### Example 1: Windsurfing speed change

A windsurfer is traveling along a straight line, and her motion is given by the velocity graph below.
Select all of the following statements that are true about the speed and acceleration of the windsurfer.
(A) Speed is increasing.
(B) Acceleration is increasing.
(C) Speed is decreasing.
(D) Acceleration is decreasing.
Options A, speed increasing, and D, acceleration decreasing, are both true.
The slope of a velocity graph is the acceleration. Since the slope of the curve is decreasing and becoming less steep this means that the acceleration is also decreasing.
It might seem counterintuitive, but the windsurfer is speeding up for this entire graph. The value of the graph, which represents the velocity, is increasing for the entire motion shown, but the amount of increase per second is getting smaller. For the first 4.5 seconds, the speed increased from 0 m/s to about 5 m/s, but for the second 4.5 seconds, the speed increased from 5 m/s to only about 7 m/s.

### Example 2: Go-kart acceleration

The motion of a go-kart is shown by the velocity vs. time graph below.
A. What was the acceleration of the go-kart at time t, equals, 4, start text, space, s, end text?
B. What was the displacement of the go-kart between t, equals, 0, start text, space, s, end text and t, equals, 7, start text, space, s, end text?

#### A. Finding the acceleration of the go-kart at $t=4\text{ s}$t, equals, 4, start text, space, s, end text

We can find the acceleration at t, equals, 4, start text, space, s, end text by finding the slope of the velocity graph at t, equals, 4, start text, space, s, end text.
start text, s, l, o, p, e, end text, equals, start fraction, start text, r, i, s, e, end text, divided by, start text, r, u, n, end text, end fraction
For our two points, we'll choose the start—3, start text, space, s, end text, comma, 6, start text, space, m, slash, s, end text—and end—7, start text, space, s, end text, comma, 0, start text, space, m, slash, s, end text—of the diagonal line as points one and two respectively. Plugging these points into the formula for slope we get
start text, s, l, o, p, e, end text, equals, start fraction, v, start subscript, 2, end subscript, minus, v, start subscript, 1, end subscript, divided by, t, start subscript, 2, end subscript, minus, t, start subscript, 1, end subscript, end fraction, equals, start fraction, 0, start text, space, m, slash, s, end text, minus, 6, start text, space, m, slash, s, end text, divided by, 7, start text, space, s, end text, minus, 3, start text, space, s, end text, end fraction, equals, start fraction, minus, 6, start text, space, m, slash, s, end text, divided by, 4, start text, space, s, end text, end fraction, equals, minus, 1, point, 5, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction
start text, a, c, c, e, l, e, r, a, t, i, o, n, end text, equals, minus, 1, point, 5, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction

#### B. Finding the displacement of the go-kart between $t=0\text{ s}$t, equals, 0, start text, space, s, end text and $t=7\text{ s}$t, equals, 7, start text, space, s, end text

We can find the displacement of the go-kart by finding the area under the velocity graph. The graph can be thought of as being a rectangle (between t, equals, 0, start text, space, s, end text and t, equals, 3, start text, space, s, end text) and a triangle (between t, equals, 3, start text, space, s, end text and t, equals, 7, start text, space, s, end text). Once we find the area of these shapes and add them, we will get the total displacement.
The area of the rectangle is found by
start text, a, r, e, a, end text, equals, h, times, w, equals, 6, start text, space, m, slash, s, end text, times, 3, start text, space, s, end text, equals, 18, start text, space, m, end text
The area of the triangle is found by
start text, a, r, e, a, end text, equals, start fraction, 1, divided by, 2, end fraction, b, h, equals, start fraction, 1, divided by, 2, end fraction, left parenthesis, 4, start text, space, s, end text, right parenthesis, left parenthesis, 6, start text, space, m, slash, s, end text, right parenthesis, equals, 12, start text, space, m, end text
Adding these two areas together gives the total displacement.
start text, t, o, t, a, l, space, a, r, e, a, end text, equals, 18, start text, space, m, end text, plus, 12, start text, space, m, end text, equals, 30, start text, space, m, end text
start text, t, o, t, a, l, space, d, i, s, p, l, a, c, e, m, e, n, t, end text, equals, 30, start text, space, m, end text