question_answer

Two cars are moving on a straight road with the one car 200m away from the other at the same speed. At a certain moment, brakes are applied to the rear car, which decelerates uniformly, travelling 20 m before coming to rest. Find the distance between the cars at the moment the rear car comes to rest.

Answer:

Let A and B be the positions of two cars such that AB = 200m C be the position of rear car after coming to rest and D be the position of first car CD be the distance between two cars after the rear car comes to rest \[CD=?\] From figure, \[CD=CB+BD=180+BD\] \[\therefore CD=180+ut\,\,(\therefore S=vt)\] ….. (1) Applying \[S=ut+\frac{1}{2}a{{t}^{2}}\]for AC, we get, \[20=ut+\frac{1}{2}a{{t}^{2}}\]  ...........(2)(\[DE=(2x)\]body is retarding) and \[v=a+at\Rightarrow 0=u-at\] \[\Rightarrow u=at\Rightarrow a=\frac{u}{t}\] Substituting value of a in (2); we get, \[20=ut-\frac{1}{2}\times \frac{u}{t}\times {{t}^{2}}\Rightarrow 20=ut-\frac{1}{2}ut,\,\Rightarrow ut=40\]                                                           …. (3) Substituting (3) in (1), we get, \[\therefore CD=180+BD=180+40=220m.\]