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

DEscription	Equation
Displacement-velocity relation.	$\begin{aligned}s &= \int v \ dt \\\hspace{5mm}\\v&= \cfrac{ds}{dt}\end{aligned}$
Displacement-acceleration relation.	$\begin{aligned}s&= \int\left(\int a \ dt\right) dt\\\hspace{5mm}\\a &= \cfrac{d^2 s}{dt^2}\end{aligned}$
Velocity-acceleration relation.	$\begin{aligned}v &= \int a \ dt \\\hspace{5mm}\\a &= \cfrac{dv}{dt}\end{aligned}$

Variable definitions

quantity name	symbol	unit name	unit
$Displacement$	$s$	$Metres$	$m$
$Velocity$	$v$	$Metres \ per\ second$	$ms^{-1}$
$Acceleration$	$a$	$Metres \ per\ second \ squared$	$ms^{-2}$
$Time$	$t$	$Seconds$	$s$

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:

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