Constant acceleration formulae

​​In a nutshell

When acceleration is constant, you can use the constant acceleration formula to find missing variables. You can use integration to derive these formulae.

Equations

Variable definitions

Deriving the formula

Take a particle which is moving in a straight line with a constant acceleration of $$a\ ms^{-2}$$. The particle has an initial velocity of $$u\ ms^{-1}$$​ and an initial displacement of $$0m$$. Find a formula for its velocity $$v$$ and its displacement $$s$$ at time $$t$$.​

Velocity can be found by integrating acceleration:

​$$v= \int{a}\ dt$$

$$v= at+c$$​​​

The initial velocity is $$u$$. When $$t=0,\ v=u$$. Substitute:​

​$$u=a(0)+c =c$$​​

Therefore:

​$$\boxed{v = u+at}$$​​

Displacement can be found by integrating velocity:

​$$s= \int{v}\ dt$$​​

Substitute the equation for velocity:

​$$s= \int{u+at}\ dt$$

​$$s= ut+\dfrac12at^2+c$$

​​​

When $$t=0, s=0$$: ​

​$$0=u(0)+\dfrac12a(0^2)+c$$

$$c=0$$​​​

Therefore:

​$$\boxed{s=ut+\dfrac12at^2}$$​​

Note: You may be asked to prove the constant acceleration formulae. You must be able to use integration to derive these formulae.

Example:

A cyclist moves in a straight line with a constant acceleration of $$4\ ms^{-2}$$. Given that his initial velocity is $$10\ ms^{-1}$$, show that his velocity at time $$t$$ is given by $$v=10+4t$$.

Velocity is found by integrating acceleration:

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

$$v=at+c$$​​​

$$a=4$$:

$$v=4t+c$$​​

At $$t=0,\ v=10$$:

$$10=4(0)+c=c$$​​

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Therefore, the cyclist's velocity at time $$t$$ is given by $$\underline{v=10+4t}$$.