1.

A sports car passing a police check post at60 km h^(-1)immediately started slowing down uniformly until its speed was40 km h^(-1). It continued to move at the same speed until it was passed by a police car1 km from the check post. The police car had started from resta t the checkpost at the same instant as the speorts car had passed the chstant as the sports car had passed the check post. The police car had moved with a constant acceleraation until it had passed sports car. Assuming that the time taken by the sports car in slowing down from 60 km h^(-1) to40 km h^(-1) was equal to the time that it travelled at constant speed before passed by the police car, find (a) the time taken by the police car to reach the sports car,(b) the speed of the policecar at the instant when it passed the sports caar, (c ) the time measured from the check post when the speeds of the two cars wer equal.

Answer» <html><body><p></p>Solution :Refer to fing . 2 (APC). 1 , wher (O) is the check post and cr is moving aong the path ` <a href="https://interviewquestions.tuteehub.com/tag/ox-584365" style="font-weight:bold;" target="_blank" title="Click to know more about OX">OX</a>` . <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_XI_V01_C02_S01_527_S01.png" width="80%"/>. <br/> The car goes from ` O to A` in time (t) with a constant retardation and ` A to B` in time (t) with a constant velocity (= 40 km h^(-1)), where ` <a href="https://interviewquestions.tuteehub.com/tag/ob-582890" style="font-weight:bold;" target="_blank" title="Click to know more about OB">OB</a> =1 km`, and at (B), the speorts car is overtaken by <a href="https://interviewquestions.tuteehub.com/tag/police-602070" style="font-weight:bold;" target="_blank" title="Click to know more about POLICE">POLICE</a> car.Therefore time taken by car to go from ` A to B` is also (t) as <a href="https://interviewquestions.tuteehub.com/tag/per-590802" style="font-weight:bold;" target="_blank" title="Click to know more about PER">PER</a> question. ltBrgt (a) Average velocity of sports car fro motion from ` O to A= ( 60 + 40 0 ///2 = 50 km H^(-1)` <br/> :. ( 50 xx t) + 40 xx t = 1 km or t = 1 /(90) h` <br/> Time taken by polece car to reach sports car is ` <br/> ` =2t = 2 = 2 xx 1/(90) = 1/(45) h = 1 /(45) xx 60 xx 60 = 80 s` <br/> (b) Let (a) be he acceleration of police car and (v) be its velocity when it overtakes the sports car. <br/> ` Average velocity ` = (0 + v)/2 = v/2` <br/> <a href="https://interviewquestions.tuteehub.com/tag/distanc-2586883" style="font-weight:bold;" target="_blank" title="Click to know more about DISTANC">DISTANC</a> e=average velocity xx time taken <br/> ` 1 km = v/2 xx 2t =vt =v xx 1/(90)` <br/> ` v=90 km h^(-1)` <br/> (c ) Taking motion of polece car from ` O to B`, we have <br/> ` v=u +at` <br/>:. ` 90 =0 = a xx (2t) =a xx ( 1//45)` <br/> or ` a= 90 xx 45 km h^(-1)` <br/> Let (b) be the retardation of sports car. Takin gmotion of sports car from ` o to A` we have <br/> ` u= 60 km h^(-1) , v= 40 km h^(-1)` , <br/> ` t= 1 /(90) h,A=b= ?` <br/> As ` v=u+ At` <br/> :. 40 =60 + b xx 1/(90) or b =- 20 =- 90 km h^(-2)` <br/> Let the velocity of the two cars be same at time ` t_1` then veloicty gained by police car in time ` t_1` = velocity gained by sports car in time` t-1)` <br/> :. ` 0 + (90 xx 45 ) t-1 =60 - (20 xx 90 0 t_1 <br/> or ` t_1 (60)/(90 xx 65) h = (60 xx 60 xx 60 )/ (90 xx 65) = 37 s` .</body></html>


Discussion

No Comment Found

Related InterviewSolutions