Variable acceleration in one dimension

​​In a nutshell

Sometimes an object will have variable acceleration. This means the acceleration of the object varies over time and can therefore be written as a function of time. These type of problems are similar to the ones which involve constant velocities or constant accelerations. 

Equations

Variable definitions

Variable acceleration in one dimension

Example 1

A particle undergoes acceleration with $$a = \left(10t + \cfrac{5}{2}\right) \ ms^{-2}$$. It is given that $$v = 2 \ m s^{-1}$$ when $$t=0 \ s.$$​

i) Calculate the velocity when $$t=10 \ s.$$​

ii) Calculate the distance travelled after $$10 \ s$$, to one decimal place.

i) Calculate the velocity when $$t=10 \ s.$$​

Use $$v = \int a \ dt$$: 

$$\begin{aligned}v &= \int a \ dt \\&= \int \left(10 t +\cfrac{5}{2}\right) \ dt \\&= \int 10 t \ dt + \int \cfrac{5}{2} \ dt \\& = \cfrac{10t^2}{2} + \dfrac52t + c \\&= 5t^2 + \dfrac52t + c\end{aligned}$$​

Given that $$v=2\ ms^{-1}$$ when $$t=0 \ s$$, you can find out the value of $$c.$$​

$$\begin{aligned}2 &= 5 \times 0^2 + \frac 52 \times 0 + c\\c &= 2\end{aligned}$$​​

Therefore, the expression for the velocity as a function of time is:

$$v(t) = 5t^2 + \frac 52 t + 2$$​​

Substitute $$t=10 \ s.$$ gives: 

$$\underline{v = 527 \ ms^{-1}}$$​​

ii) Calculate the distance travelled after $$t=10 \ s$$, to one decimal place.

Use $$s = \int v \ dt$$:

$$\begin{aligned}s &= \int_{t_0}^{t} v \ dt \\&= \int_{t_0=0\ s}^{t= 10 \ s} (5t^2 +\frac 5 2t + 2)\ dt \\&=\left[ \cfrac{5t^3}{3} + \cfrac{5t^2}{4} + 2t\right]_{0}^{10} \\&= \cfrac{5000}{3} + \cfrac{500}{4} + 20 \\&= 1811.667.. \ m\end{aligned}$$​​

Therefore, $$\underline{s = 1810 \ m \ (3 \ s.f.)}$$.

Example 2

A particle with $$m=10 \ kg$$ is moving along the positive $$x$$​-axis. At a time $$t$$, the displacement is $$s= \left(2t^{3/2} +\cfrac{ e^{-2t}}{3}\right) m$$. 

i) Calculate the velocity when $$t = 1.5 \ s$$ to one decimal place.

ii) Given that the particle is acted on by a force $$F$$, calculate the value of $$F$$ when $$t=10 \ s$$ to one decimal place.

i) Calculate the velocity when $$t=1.5 \ s$$ to one decimal place.

Use $$v = \cfrac{ds}{dt}$$ : 

$$\begin{aligned}v &= \cfrac{ds}{dt} \\&= \cfrac{d}{dt}\left(2t^{3/2} + \cfrac{e^{-2t}}{3}\right)\\& = 3\sqrt{t} - \cfrac{2e^{-2t}}{3}\end{aligned}$$​​

Substitute $$t=1.5 \ s$$ and you obtain:​

​$$\underline{v= 3.64 \ ms^{-1} \ (3 \ s.f.)}$$​​

ii) Given that the particle is acted on by a force $$F$$, calculate the value of $$F$$ when $$t=10 \ s$$ to one decimal place.

By definition $$F = m\times a$$. Therefore, first you need to calculate the acceleration:​

​$$\begin{aligned}a&=\cfrac{dv}{dt} \\&= \cfrac{d}{dt}\left(3\sqrt{t} - \cfrac{2e^{-2t}}{3}\right)\\&= \cfrac{4e^{-2t}}{3} + \cfrac{3}{2\sqrt{t}}\end{aligned}$$​​

Substituting $$t = 10 \ s$$ you have: 

$$a = 0.47434165... \ ms^{-2}$$​​

Thus, the force is: 

​$$\underline{F = 4.74 \ N \ (3 \ s.f.)}$$​​