Velocity-time graphs

If an object moves along a straight line, its motion can be represented by a velocity-time graph. The gradient of the line is equal to the acceleration of the object.

A velocity/time graph. Graph with four distinct sections. All lines are straight.

The table shows what each section of the graph represents:

Section of graph	Gradient	Velocity	Acceleration
A	postive	increasing	postive
B	zero	constant	zero
C	negative	decreasing	negative
D	zero	stationary (at rest)	zero

Section of graph	A
Gradient	postive
Velocity	increasing
Acceleration	postive

Section of graph	B
Gradient	zero
Velocity	constant
Acceleration	zero

Section of graph	C
Gradient	negative
Velocity	decreasing
Acceleration	negative

Section of graph	D
Gradient	zero
Velocity	stationary (at rest)
Acceleration	zero

Example

Describe the motion of the vehicle in the graph at the three stages of its journey.

Between 0 s and 4 s the vehicle is accelerating.

\(acceleration = \frac{change~in~velocity}{time~taken}\)

This means that: \(8 \div 2 = 4 m/s^2\)

Between 4 s and 7 s the vehicle has a constant velocity of 8 m/s.

Between 7 s and 10 s, the vehicle is decelerating so: \(-8 \div 3 = -2.67~m/s^2\)

Calculating displacement - Higher

The of an object can be calculated from the area under a velocity-time graph.

The area under the graph can be calculated by:

using geometry (if the lines are straight)
counting the squares beneath the line (particularly if the lines are curved)

Example

Calculate the total displacement of the object whose motion is represented by the velocity-time graph below.

The y axis shows velocity in metres per second and the x axis time in seconds. The object increases its velocity from 0 metres per second to 8 metres per second in 4 seconds.

The displacement can be found by calculating the total area of the shaded sections below the line.

Find the area of the triangle:

\(\frac{1}{2} \times base \times height\)

\(\frac{1}{2} \times 4 \times 8 = 16~m^{2}\)

Find the area of the rectangle:

base × height

(10 - 4) × 8 = 48 m²

Add the areas together to find the total displacement:

(16 + 48) = 64 m

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Motion - OCR GatewayVelocity-time graphs