1.

The brakes applied to a car produce an acceleration of `6 m//s^(2)` in the opposite direction to the motion . If the car takes ` 2 s` to stop after the application of brakes , calculate the distance it travels during this time.

Answer» We have been given
`a=-6 m s^(-2) , t=2 s` and `v=0 m s^(-1)`
From Eq. (8.5) we know that
`v=ut+at`
`0=u+(-6 m s^(-2))xx2 s`
or `u=12 m s^(-1)` .
From Eq. (8.6) we get
`s=ut+(1)/(2) at^(2)`
`=(12 m s^(-1))xx(2 s)+(1)/(2)(-6 m s^(-2))(2 s)^(2)`
`=24 m -12 m `
=12 m
Thus, the car will move 12 m before itstops after the application of brakes.Can you now appreciate why driversare cautioned to maintain somedistance between vehicles whiletravelling on the road?


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