Using integration to find velocity and displacement

​​In a nutshell

Integration is the reverse process of differentiation. You can integrate acceleration with respect to time to find velocity. You can integrate velocity with respect to time to find displacement.

Equations

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Variable definitions

Velocity

To find velocity, integrate acceleration with respect to time. To integrate, you do the reverse process of differentiation. However, when integrating you must also include a constant $$c$$ which you must work out using information from the question.​

Example 1

A car is accelerating down a road. The formula for the acceleration at time $$t$$ is $$a=4t+6$$. At time $$t=2$$, the velocity of the car is $$4ms^{-1}$$. Find a formula for the velocity of the car at time $$t$$.

Integrate the given formula for acceleration:

$$v=\int{a}\ dt$$​​

$$v=\int{4t+6}\ dt$$

​$$v=2t^2+6t+c$$​

At $$t=2,\ v=4$$. Substitute this into the formula to find $$c$$​:

$$4=2(2^2)+6(2)+c$$​​

$$4=8+12+c$$

$$c=4-20=-16$$

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Use this value for $$c$$ and substitute into the formula:

The formula for velocity at time $$t$$ is $$\underline{v=2t^2+6t-16}$$.

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Displacement

To find displacement, integrate velocity with respect to time. 

Example 2

A cyclist is travelling away from home. At time $$t=0, \ s=0$$. The cyclist's velocity for the first $$5s$$​ is modelled by $$v=3t^2-6t+3$$. Find the cyclist's distance from home when $$t=3$$

Integrate the formula for velocity:

$$s=\int{v}\ dt$$

$$s =\int{3t^2-6t+3}\ dt$$

$$s=t^3-3t^2+3t+c$$​​​​

At $$t=0,\ s=0$$. Substitute:​

​$$0=0^3-3(0^2)+3(0)+c$$

$$c=0$$

Therefore:

​$$s= t^3-3t^2+3t$$​​

Substitute $$t=3$$:

​$$s=3^3-3(3^2)+3(3)$$

$$s=27-27+9=9$$​​​

The cyclist is $$\underline{9m}$$ away from home at $$3s$$.