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In a car race, car A takes time t less than car B and passes the finishing point with a velocity v more than the velocity with which car B passes the point. Assuming that the cars start from rest and travel with constant accelerations a_(1)anda_(2), show that (v)/(t)=sqrt(a_(1)a_(2)).

Answer» <html><body><p></p>Solution :Let s be the distance covered by each car. Let the times taken by the two <a href="https://interviewquestions.tuteehub.com/tag/cars-909967" style="font-weight:bold;" target="_blank" title="Click to know more about CARS">CARS</a> to <a href="https://interviewquestions.tuteehub.com/tag/complete-423576" style="font-weight:bold;" target="_blank" title="Click to know more about COMPLETE">COMPLETE</a> the journey be `t_(1)andt_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)`, and their velocities at the finishing point be `v_(1) and v_(2)` respectively. <br/> According to the given <a href="https://interviewquestions.tuteehub.com/tag/problem-25530" style="font-weight:bold;" target="_blank" title="Click to know more about PROBLEM">PROBLEM</a>, <br/> `v_(1)-v_(2)=<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> and t_(2)-t_(1)=t` <br/> Now, `(v)/(t)=(v_(1)-v_(2))/(t_(2)-t_(1))=(sqrt(2a_(1)s)-sqrt(2a_(2)s))/(sqrt((2s)/(a_(2)))-sqrt((2s)/(a_(1))))=(sqrt(a_(1))-sqrt(a_(2)))/(sqrt((1)/(a_(2)))-sqrt((1)/(a_(1))))` <br/> `therefore (V)/(t)=sqrt(a_(1)a_(2))`</body></html>


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