Answers

2016-02-19T20:18:55+05:30
Answer :  \frac{1}{16}

Solution:

By Newtons third law of motion
v^{2}-u^{2}=2as

here the acceleration a is negative since it is retardation, displacement s is the stopping distance d, u is the initial velocity and v is the final velocity.
Since the objects stops after the motion, v=0
That is
  -u^{2} = -2ad
   ⇒d= \frac{u}{2a}

Lets take d_{1} as the distance traveled by the car with velocity u and d_{2} as the distance traveled by the car with velocity  4u

The retardation for both the cars is -a
   ⇒ \frac{d_{1}}{d_{2}}=  \frac{u^{2}}{2a} \frac{2a}{(4u)^{2}}
   ⇒ \frac{d_{1}}{d_{2}}=\frac{1}{16}
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