A retarding force is applied to stop a motor car . if the speed of motor car is doubled, then what will be the distance covered before stopping the same
Answers
Answered by
32
Stopping Distance:
s=v^2−u^2/2a
=0−u^2/2(−F/m)
=mu^2/2F
As m and F are constants,
sαu^2
s2/s1=(u2/u1)^2
=(2u/u)^2
s2=4s1
Ans: Stopping distance increases to 4 times.
s=v^2−u^2/2a
=0−u^2/2(−F/m)
=mu^2/2F
As m and F are constants,
sαu^2
s2/s1=(u2/u1)^2
=(2u/u)^2
s2=4s1
Ans: Stopping distance increases to 4 times.
Answered by
2
Clearly the motor car Will cover 4 times more distance
Similar questions