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Constant acceleration

Constant acceleration: Deriving formulae

Constant acceleration: Deriving formulae

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Explainer Video

Tutor: Mohammed

Summary

Constant acceleration: Deriving formulae

​​In a nutshell

Using the velocity-time graph, you can derive more formulae to work out the motion of an object with constant acceleration. 


Equations


description

equation

Velocity of an object with constant acceleration.
​v=u+atv=u+atv=u+at​​
Displacement of an object over a certain time.
​s=(u+v2)ts=\bigg(\dfrac{u+v}{2}\bigg)ts=(2u+v​)t​​
Acceleration of an object.
​a=v−uta=\dfrac{v-u}{t}a=tv−u​​​


Variable definitions


quantity name

symbol

unit name

unit

DisplacementDisplacementDisplacement​​
​sss​​
​MetreMetreMetre​​
​mmm​​
​Initial velocityInitial\ velocityInitial velocity​​
​uuu​​
​Metres per secondMetres\ per\ secondMetres per second​​
​ms−1ms^{-1}ms−1​​
​Final velocityFinal\ velocityFinal velocity​​
​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


​

Deriving formulae

You need to know how to derive two formulae from a velocity-time graph.

Maths; Constant acceleration; KS5 Year 12; Constant acceleration: Deriving formulae


The acceleration of an object can be worked out using the gradient of a velocity-time graph: 


a=v−uta=\dfrac{v-u}{t}a=tv−u​​​


Rearrange this to make vvv the subject:

​at=v−uat=v-uat=v−u

​​

v=u+at\boxed{v = u+at}v=u+at​​​


The displacement of an object can be worked out by the area under the graph. You can work out the area using the formula for the area of a trapezium:


​Area of a trapezium=(a+b2)hArea\ of\ a\ trapezium= \bigg(\dfrac{a+b}{2}\bigg)hArea of a trapezium=(2a+b​)h


In the velocity-time graph:​

a=u−0=ua=u-0=ua=u−0=u, b=v−0=vb=v-0=vb=v−0=v, h=th=th=t

​​

Substitute these values into the formula for the area of a trapezium:


​s=(u+v2)t\boxed{s=\bigg(\dfrac{u+v}{2}\bigg)t}s=(2u+v​)t​​​

 

These formulae can be rearranged to make each variable the subject, and then applied to objects in motion.


Example

An object starts at point A with a velocity of 2 ms−12\ ms^{-1}2 ms−1 and gets to point B after 202020​ seconds, reaching a velocity of 10 ms−110\ ms^{-1}10 ms−1. Work out:

a: The total displacement between points A and B.

b: The acceleration of the object between these points.


Start by identifying the known variables:

u=2 ms−1u=2\ ms^{-1}u=2 ms−1, v=10 ms−1v=10\ ms^{-1}v=10 ms−1, t=20 st=20\ st=20 s


a: Formula for displacement: 

s=(u+v2)ts=\bigg(\dfrac{u+v}{2}\bigg)ts=(2u+v​)t

​​

Substitute the known variables:

s=(2+102)20s=\Big( \dfrac{2+10}{2}\Big)20s=(22+10​)20​​


s=(122)20=6×20=120s = \bigg(\dfrac{12}{2}\bigg)20 = 6\times 20 = 120s=(212​)20=6×20=120​​


The displacement of the object is 120 m‾\underline{ 120 \ m}120 m​.


b: Formula for acceleration:

a=v−uta=\dfrac{v-u}{t}a=tv−u​


Substitute the known variables:

a=10−220=820=410a=\dfrac{10-2}{20}=\dfrac{8}{20}=\dfrac{4}{10}a=2010−2​=208​=104​​​


The acceleration of the object is 0.4 m⋅s−2‾\underline{0.4 \ m\cdot s^{-2}}0.4 m⋅s−2​.


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Speed, velocity and acceleration

Unit 1

Speed, velocity and acceleration

Acceleration and velocity-time graphs

Unit 2

Acceleration and velocity-time graphs

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Constant acceleration: Deriving formulae

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Constant acceleration: Deriving formulae

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FAQs - Frequently Asked Questions

What is the formula for final velocity when there is constant acceleration?

v=u+at.

How do I work out acceleration from a velocity-time graph?

The acceleration of an object can be worked out using the gradient of a velocity-time graph.

How do I work out displacement from a velocity-time graph?

The displacement of an object can be worked out by the area under the velocity-time graph.

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