The correct option is D 10tan−1(10)−12ln|1+102|
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, dvdt=a=tan−1(t).
To find v, we will integrate acceleration.
We have dv=tan−1(t)dt
⇒v(t)=∫tan−1(t)dt
We solved this integral before. To solve that we apply integration by partial fraction in the following way
∫tan−1(t)dt=∫(tan−1(t)×1)dt
=tan−1(t)∫1dt−∫[ddt(tan−1(t)∫1dt]dt
=tan−1(t)t−∫11+t2tdt
=t⋅tan−1(t)−12∫2t1+t2dt
In the second term, if substitute,
u=1+t2⇒du=2tdt which is the numerator.
So we get
∫tan−1(t)dt=t.tan−1(t)−12∫duu
=t⋅tan−1(t)−12ln|u|
Substituting back u=1+t2
∫tan−1(t)dt=t tan−1(t)−12ln|1+t2|+c
⇒v(t)=t tan−1(t)−12ln|1+t2|+c
We have v(0)=0
⇒c=0
Velocity at t =10,
v(10)=10tan−1(10)−12ln|1+100|)