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Speed of two identical cars are u and 4u at a specific instant,the ratio of the respective distance in which the two cars are stopped from that instant by

same breaking force-


Answer :  \frac{1}{16}


By Newtons third law of motion

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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