A particle moves along a straight line its velocity dipends on time as `v=4t-t^(2)`. Then for first `5 s`:

Options

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

C, D

Explanation

`"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)`