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Question

A car moves with a variable acceleration given by a=tan1(t) where t is the time in seconds. Find the velocity of the car after 10 seconds, if it was initially at rest

A
10tan1(10)12ln|102|
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B
10tan1(1+102)12ln|102|
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C
tan1(10)12ln|1+102|
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D
10tan1(10)12ln|1+102|
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Solution

The correct option is C 10tan1(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=tan1(t).
To find v, we will integrate acceleration.
We have dv=tan1(t)dt
v(t)=tan1(t)dt
We solved this integral before. To solve that we apply integration by partial fraction in the following way
tan1(t)dt=(tan1(t)×1)dt

=tan1(t)1dt[ddt(tan1(t)1dt]dt

=tan1(t)t11+t2tdt

=ttan1(t)122t1+t2dt
In the second term, if substitute,
u=1+t2du=2tdt which is the numerator.
So we get

tan1(t)dt=t.tan1(t)12duu

=ttan1(t)12ln|u|
Substituting back u=1+t2
tan1(t)dt=t tan1(t)12ln|1+t2|+c
v(t)=t tan1(t)12ln|1+t2|+c
We have v(0)=0
c=0
Velocity at t =10,
v(10)=10tan1(10)12ln|1+100|)


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