Question

A car moves with a variable acceleration given by $a={\tan }^{-1}\left(t\right)$ where t is the time in seconds. Find the velocity of the car after 10 seconds, if it was initially at rest

• A. $10{\tan }^{-1}\left(10\right)-\frac{1}{2}\ln |{10}^2|$
• B. $10{\tan }^{-1}(1+{10}^2)-\frac{1}{2}\ln |{10}^2|$
• C. ${\tan }^{-1}\left(10\right)-\frac{1}{2}\ln |1+{10}^2|$
• D. $10{\tan }^{-1}\left(10\right)-\frac{1}{2}\ln |1+{10}^2|$

Answer 1

The correct option is D $10{\tan }^{-1}\left(10\right)-\frac{1}{2}\ln |1+{10}^2|$

We are given that car was initially at rest. So the velocity, v, at time t = 0 is zero .
v(0) = 0
We know that acceleration is the rate of change of velocity
Or, $\frac{dv}{dt}=a={\tan }^{-1}\left(t\right)$.
To find v, we will integrate acceleration.
We have dv=${\tan }^{-1}\left(t\right)dt$
$\Rightarrow v\left(t\right)=\int {\tan }^{-1}\left(t\right)dt$
We solved this integral before. To solve that we apply integration by partial fraction in the following way
$\int {\tan }^{-1}\left(t\right)dt=\int \left({\tan }^{-1}\right(t)\times 1)dt$

$={\tan }^{-1}\left(t\right)\int 1dt-\int \left[\frac{d}{dt}\right({\tan }^{-1}\left(t\right)\int 1dt]dt$

$={\tan }^{-1}\left(t\right)t-\int \frac{1}{1+t^2}tdt$

$=t\cdot {\tan }^{-1}\left(t\right)-\frac{1}{2}\int \frac{2t}{1+t^2}dt$
In the second term, if substitute,
$u=1+t^2\Rightarrow du=2tdt$ which is the numerator.
So we get
$\int {\tan }^{-1}\left(t\right)dt=t.{\tan }^{-1}\left(t\right)-\frac{1}{2}\int \frac{du}{u}$

$=t\cdot {\tan }^{-1}\left(t\right)-\frac{1}{2}\ln |u|$
Substituting back $u=1+t^2$
$\int {\tan }^{-1}\left(t\right)dt=t{\tan }^{-1}\left(t\right)-\frac{1}{2}\ln |1+t^2|+c$
$\Rightarrow v\left(t\right)=t{\tan }^{-1}\left(t\right)-\frac{1}{2}\ln |1+t^2|+c$
We have v(0)=0
$\Rightarrow c=0$
Velocity at t =10,
$v\left(10\right)=10{\tan }^{-1}\left(10\right)-\frac{1}{2}\ln |1+100|)$