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Variable acceleration

Using integration to find velocity and displacement

Using integration to find velocity and displacement

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Explainer Video

Tutor: Mohammed

Summary

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

DESCRIPTION

EQUATION

Integrating acceleration to find velocity.
​v=∫a dtv = \int{a}\ dtv=∫a dt​​
Integrating velocity to find displacement.
​s=∫v dts = \int{v}\ dts=∫v dt​​

​

Variable definitions

QUANTITY NAME

SYMBOL

UNIT NAME

UNIT

DisplacementDisplacementDisplacement​​
​sss​​
​MetreMetreMetre​​
​mmm​​
​VelocityVelocityVelocity​​
​vvv​​
​Metres per secondMetres\ per\ secondMetres per second​​
​ms−1ms^{-1}ms−1​​
​AccelerationAccelerationAcceleration​​
​aaa​​
​Metres per second squaredMetres\ per\ second\ squaredMetres per second squared​​
​ms−2ms^{-2}ms−2​​
​TimeTimeTime​​
​ttt​​
​SecondsSecondsSeconds​​
​sss



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 ccc 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 ttt is a=4t+6a=4t+6a=4t+6. At time t=2t=2t=2, the velocity of the car is 4ms−14ms^{-1}4ms−1. Find a formula for the velocity of the car at time ttt.


Integrate the given formula for acceleration:

v=∫a dtv=\int{a}\ dtv=∫a dt​​

v=∫4t+6 dtv=\int{4t+6}\ dtv=∫4t+6 dt

​v=2t2+6t+cv=2t^2+6t+cv=2t2+6t+c​

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

4=2(22)+6(2)+c4=2(2^2)+6(2)+c4=2(22)+6(2)+c​​

4=8+12+c4=8+12+c4=8+12+c

c=4−20=−16c=4-20=-16c=4−20=−16

​​​

Use this value for ccc and substitute into the formula:


The formula for velocity at time ttt is v=2t2+6t−16‾\underline{v=2t^2+6t-16}v=2t2+6t−16​.

​


Displacement

To find displacement, integrate velocity with respect to time. 


Example 2

A cyclist is travelling away from home. At time t=0, s=0t=0, \ s=0t=0, s=0. The cyclist's velocity for the first 5s5s5s​ is modelled by v=3t2−6t+3v=3t^2-6t+3v=3t2−6t+3. Find the cyclist's distance from home when t=3t=3t=3


Integrate the formula for velocity:

s=∫v dts=\int{v}\ dts=∫v dt

s=∫3t2−6t+3 dts =\int{3t^2-6t+3}\ dts=∫3t2−6t+3 dt

s=t3−3t2+3t+cs=t^3-3t^2+3t+cs=t3−3t2+3t+c​​​​

At t=0, s=0t=0,\ s=0t=0, s=0. Substitute:​

​0=03−3(02)+3(0)+c0=0^3-3(0^2)+3(0)+c0=03−3(02)+3(0)+c

c=0c=0c=0

Therefore:

​s=t3−3t2+3ts= t^3-3t^2+3ts=t3−3t2+3t​​

Substitute t=3t=3t=3:

​s=33−3(32)+3(3)s=3^3-3(3^2)+3(3)s=33−3(32)+3(3)

s=27−27+9=9s=27-27+9=9s=27−27+9=9​​​


The cyclist is 9m‾\underline{9m}9m​ away from home at 3s3s3s.


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FAQs - Frequently Asked Questions

What is integration?

Integration is the reverse process of differentiation.

What do you integrate to work out displacement?

You can integrate velocity with respect to time to find displacement.

What do you integrate to work out velocity?

You can integrate acceleration with respect to time to find velocity.

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