Functions of time: Displacement, velocity, acceleration

​​In a nutshell

When an object has variable acceleration it means the acceleration is changing over time. The displacement, velocity and acceleration can be modelled as functions of time.

Acceleration on a velocity-time graph

A changing acceleration can be shown on a velocity-time graph. Acceleration is the rate of change of velocity, and can be found by working out the gradient of the curve. An increasing acceleration is represented by an increasing gradient, and a decreasing acceleration is represented by a decreasing gradient. 

Displacement, velocity and acceleration can be modelled as functions in terms of time. 

Displacement as a function of time

Displacement can be modelled as a function of time. These functions can be drawn out on a graph to help visualise the scenario. The displacement function can be used to work out the displacement at any given value for time, and also to identify times when there is no displacement - on a graph this is when the function crosses the $$x$$​-axis.

Example 1

A ball moves in a straight line. Its movement can be modelled by $$s=3t^3-8t$$, where $$t>0$$. Find:

a: The displacement of the ball at $$4$$ seconds.

b: The time it takes for the ball to return to its origin.

a: At $$4$$ seconds, $$t=4$$. Substitute into the formula for displacement:

$$s=3t^3-8t$$

$$s=3(4^3)-8(4)$$

​$$s=192-32=160\ m$$​​

The displacement of the ball at four seconds is $$\underline{160\ m}$$.

b: At the origin, $$s=0$$. Therefore:

$$3t^3-8t=0$$

Factorise:

​$$t(3t^2-8)=0$$

​​

For this to be true, either $$t=0$$ or $$3t^2-8=0$$​. The time when the ball sets off is $$0$$, therefore solve $$3t^2-8=0$$:

$$3t^2-8=0$$​​

$$3t^2=8$$

$$t^2=\dfrac83$$​​​

$$t=\pm \sqrt{\dfrac83} = \pm 1.632993162$$​​

The question says the model is only valid for $$t>0$$, therefore ignore the negative:​

The ball takes $$\underline{1.64\ s}$$ to return to its origin.

Velocity as a function of time

Velocity can also be modelled as a function of time. The function can be used to work out velocity at any given time, and to identify times when an object is stationary.

Example 2

The velocity of an object is given by $$v=2t^2-26t+24$$. Find the times when the object is at rest.

The object is at rest when $$v=0$$. Therefore:

$$2t^2-26t+24=0$$​​

Solve the quadratic:

​$$t^2-13t+12=0$$

$$(t-1)(t-12)=0$$

$$t=1,\ t=12$$​​​

The object is at rest at $$\underline{1\ s}$$ and $$\underline{12\ s}$$.​