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A car starts from rest and moves with uniform acceleration 'a'. At the same instant from the same point a bike crosses with a uniform velocity 'u'. When and where will they meet? What is the velocity of car with respect to the bike at the time of meeting?

Answer» <html><body><p></p>Solution :Let the car and bike will meet after coving a <a href="https://interviewquestions.tuteehub.com/tag/distance-116" style="font-weight:bold;" target="_blank" title="Click to know more about DISTANCE">DISTANCE</a> S in a time .t. <a href="https://interviewquestions.tuteehub.com/tag/sec-1197209" style="font-weight:bold;" target="_blank" title="Click to know more about SEC">SEC</a>. <br/> For car <br/> `S=(1)/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)at^(2)……(1)` <br/> for bike <br/> `s=<a href="https://interviewquestions.tuteehub.com/tag/ut-718961" style="font-weight:bold;" target="_blank" title="Click to know more about UT">UT</a>……(2)` <br/> Since they cover same distance <br/> `ut=(1)/(2)at^(2)impliest=(2u)/(a)` <br/> The car and bike will meet after a time of `(2u)/(a)` sec. <br/> The <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> of car after .t. sec is <br/> `V_(car)=at=a((2u)/(a))=2u` <br/> The velocity of car w.r.t bike is `V_(rel)=2u-u=u`.</body></html>


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