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,
\[\therefore CD=180+BD=180+40=220m.\]
You need to login to perform this action.
You will be redirected in
3 sec