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Two bodies start moving in the same straight line at the same instant of time from the same origin. The first body moves with a constant velocity of 40m//s, and the second starts from rest with a constant acceleration of 4m//s^(2). Find the time that elapses before the second catches the first body. Find also the greatest distance between then prior to it and the time at which this occurs.

Answer» <html><body><p></p>Solution :When the <a href="https://interviewquestions.tuteehub.com/tag/second-1197322" style="font-weight:bold;" target="_blank" title="Click to know more about SECOND">SECOND</a> body catches the first, the <a href="https://interviewquestions.tuteehub.com/tag/distance-116" style="font-weight:bold;" target="_blank" title="Click to know more about DISTANCE">DISTANCE</a> travelled by each is the same. <br/> `therefore <a href="https://interviewquestions.tuteehub.com/tag/40t-315934" style="font-weight:bold;" target="_blank" title="Click to know more about 40T">40T</a>=(1)/(2)(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)t^(2)ort=20S` <br/> Now, the distance s between the two bodies at any time t is `s=ut-(1)/(2)at^(2)` <br/> For s to be maximum, `(ds)/(dt)=0oru-at=0` <br/> or `t=(u)/(a)=(40)/(4)=<a href="https://interviewquestions.tuteehub.com/tag/10s-267218" style="font-weight:bold;" target="_blank" title="Click to know more about 10S">10S</a>` <br/> Maximum Distance <br/> `=40xx10-(1)/(2)xx4xx(10)^(2)=400-200=200m`</body></html>


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