The correct option is C 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
To find v, we will integrate acceleration.
We have dv=tan−1(t)dt
We solved this integral before. To solve that we apply integration by partial fraction in the following way
In the second term, if substitute,
u=1+t2⇒du=2tdt which is the numerator.
So we get
Substituting back u=1+t2
We have v(0)=0
Velocity at t =10,