Question

# A car moves with a variable acceleration given by a=tanâˆ’1(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 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|)

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