Question

two cars are moving on a straight road with one car200 m away from the other moving at the same speed . at acertain moment brakes are applied to the rear car which decelerates uniformly traveling 20m be before coming to rest find the distance between the cars at the moment when the rear car comes to rest

Answer 1

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.

A-----C---------B---------------D

CB = 180m

AC = 20m

AB = 200m

CD=? From figure, CD=CB+BD=180+BD

CD=180+ut (S=velocity x time) .......(1)

Applying S=ut+1/2at2 for AC, we get 20=ut+1/2at2 ......(2) (DE=(2x)body is retarding) and

v=u+at

=>0=u−at

=>u=at

=>a=u/t

Substituting value of a in (2)we get,

20=ut−1/2×(u/t)×t2

=>20=ut−1/2ut

=>ut=40 .......... (3)

Substituting (3) in (1) we get,

CD=180+BD=180+40=220m.

Regards,

Answer 2

$ut + 1 \div2 {at}^{2}$

Answer:

220

Explanation

ab=200m

ac=20 m

cb=180 m

cd=?

cd=180+ut-------(1)

by using given formula mentioned above

s=1/2at^2

20 =1/2at^2-------(2)

v=u+at

0=u-at

u=at

a=u/t-------(3)

substitute 3 in the place of a in equation 2

20=ut-1/2(u/t)t^2

ut=40-------(4)

substitute 4 in equation 1

cd=180+ut

cd =180+40

CD=220