A particle moves along a straight line its velocity dipends on time as `v=4t-t^(2)`. Then for first `5 s`:A. Average velocity is `25//3 ms^(-1)`B. average speed is `10ms^(-1)`C. Average velocity is `5//3 ms^(-1)`D. acceleration is `4ms^(-2)` at `t=0`

A particle moves along a straight line its velocity dipends on time as

1.	A particle moves along a straight line its velocity dipends on time as `v=4t-t^(2)`. Then for first `5 s`:A. Average velocity is `25//3 ms^(-1)`B. average speed is `10ms^(-1)`C. Average velocity is `5//3 ms^(-1)`D. acceleration is `4ms^(-2)` at `t=0`
Answer» Correct Answer - C::D `"Average velocity"=vecv=(int_(0)^(5)vdt)/(int_(0)^(5)dt)=(int_(0)^(5)(4t-t^(2))dt)/(int_(0)^(5)dt)` `=([2t^(2)-(t^(3))/(3)]_(0))/(5)=(50-(125)/(3))/(5)=(25)/(3xx5)=(5)/(3)` For average speed let us put `v=0` which gives `t=0` and `t=4s`. `because"average speed"=(\|int_(0)^(4)vdt\|int_(4)^(5)vdt\|)/(int_(0)^(5)dt)=(\|int_(0)^(4)(4t-t^(2))dt\|+\|int_(4)^(5)vdt\|)/(5)` `=([2t^(2)-(t^(3))/(3)]_(0)^(4)+[2t^(2)-(t^(3))/(3)]_(4)^(5))/(5)` `=(\|[2t^(2)-(t^(3))/(3)]_(0)^(4)\|+\|[2t^(2)-(t^(3))/(3)]_(4)^(5)\|)/(5)=(13)/(5)ms^(-1)` For acceleration: `a=(dv)/(dt)=(d)/(dt)(4t-t^(2))=4-2t` At `t=0,a=4ms^(-2)`

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A particle moves along a straight line its velocity dipends on time as `v=4t-t^(2)`. Then for first `5 s`:A. Average velocity is `25//3 ms^(-1)`B. average speed is `10ms^(-1)`C. Average velocity is `5//3 ms^(-1)`D. acceleration is `4ms^(-2)` at `t=0`

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