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A partical is moving in a straight line such that `s=t^(3)-3t^(2)+2`, where `s` is the displacement in meters and `t` is in seconds. Find the (a) velocity at `t=2s`, (b) acceleration at `t=3s`, ( c) velocity when acceleration is zero and (d) acceleration when velocity is zero. |
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Answer» `s=t^(3)-3t^(2)+3t+2` `v=(ds)/(dt)=3t^(2)-6t+3` `a=(dv)/(dt)=6t-6` (a) At `t=2 s, v=3(2)^(2)-6(2)+3=3 m//s ` (b)At` t=1 s, a=6(3)-6=12 m//s^(2)` (c ) Acceleration `a=0implies6t-6=0impliest=1 s ` `At t=1 s, v=3(1)^(2)-6(1)+3=0` (d) When velocity is zero, `v=3t^(2)-6t+3=3(t^(2)=2t+1)=0` `(t-1)^(2)=0impliest=1 s ` At `t=1 s, a=6(1)-6=0` |
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