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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 2s to stop after the application of brakes, calculate the distance it travels during this time. |
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Answer» Solution :We have ben GIVEN `a=-6ms^(-2),t=2s` and `v=0ms^(-1)` From :EQ we know that v = u + at. `0=u+(-6ms^(-2)xx2s` or u=12 `ms^(-1)` From Eq. we get `s=ut+1/2at^(2)` `=(12ms^(-1))xx(2s)+1/2(-6ms^(-2))(2s)^(2)` = 24 m - 12 m `=12M` THUS, the car will move 12 m before it stops after the application of brakes. Can you now appreciate why drivers are cautioned to maintain some distance between vehicles while travelling on the road? |
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