InterviewSolution
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InterviewSolution
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The displacement of a particla moving in straight line is given by `s=t^(4)+2t^(3)+3t^(2)+4`, where s is in meters and t is in seconds. Find the (a) velocity at `t=1 s`, (b) acceleration at `t=2 s`, (c ) average velocity during time interval `t=0` to `t=2 s` and (d) average acceleration during time interval `t=0` to `t=1 s`. |
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Answer» `s=t^(4)+2t^(3)+3t^(2)+4` `v=(ds)/(dt)=4t^(3)+6t^(2)+6t` `a=(dv)/(dt)=12t^(2)+12t+6` (a) `At t=1 s, v=4(1)^(3)+6(1)^(2)+6(1)=16 m//s` (b) `At t=2 s, a =12(2)^(2)+12(2)+6=78 m//s^(2)` (c ) `At t_(1)=0, s_(1)=4 m` At `t_(2)=2 s, s_(2)=(2)^(4)+2(2)^(3)+3(2)^(2)+4=48 m` `bar(v)=(Deltas)/(Deltat)=(s_(2)-s_(1))/(t_(2)-t_(1))=(48-4)/(2-0)=22 m//s` `At t_(1)=0, v_(1)=0` At `t_(2)=1 s, v_(2)=4(1)^(3)+6(1)^(2)+6(1)=16 m//s` `bar(a)=(Deltav)/(Deltat)=(v_(2)-v_(1))/(t_(2)-t_(1))=(16-0)/(1-0)=16 m//s` |
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