A car is moving on a circular path of radius 600 m such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete first quarter of revolution, if it is moving with an initial speed of 54 km/hr is $t(1-e^{-\pi/2})s$. The value of t is ____________.
Answer (integer)
40
Solution
<p>$${{dv} \over {dt}} = {{{v^2}} \over R} \Rightarrow {{{v^2}} \over R} = v{{dv} \over {ds}}$$</p>
<p>$$ \Rightarrow {{dv} \over v} = {{ds} \over R} \Rightarrow \left. {\ln v} \right|_{15}^v = {s \over R}$$</p>
<p>$$ \Rightarrow v = 15{e^{\Delta /R}} = {{ds} \over {dt}} \Rightarrow dt = {1 \over {15}}{e^{ - \Delta /R}}ds$$</p>
<p>$\Delta t = {R \over {15}}[1 - {e^{ - \Delta /R}}]$</p>
<p>$= 40[1 - {e^{ - \pi /2}}]$ seconds</p>
<p>$\Rightarrow t = 40$</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Torque and Angular Momentum
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