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 velcotu is `25//3 m s^(-1)`B. Average speed is `10 m s^(-1)`.C. Average velcotu is `5//3 m s^(-1)`D. Acceleration is `4 m s^(-2)` at `t=o`

Answer» Correct Answer - C::D
Average velocity.
`vec v = (int _(0)^(5)v dt)/(int_(0)^(5) dt) = (int_(0)^(5) (4t-t^(2)) dt)/(int_(0)^(5) dt)`
`=[2t^(2) -t^(3)/3]_(0)^(3)/5 =(50 -(125)/3 )/5 =(25)/(3xx5) =5/3`
For average speed, let us put `v=0`, which gives `t=0 and `t =4` s`
For average speed, let us put `v=0`, which gives` t=0 and t=4 s`
becuase Average speed `=(|underset(0)overset(4)(int)v dt|+|underset(0)overset(5)(int)v dt|)/(underset(0)overset(5)(int)dt)=(|underset(0)overset(4)(int)4t-t^(3)dt|+|underset(4)overset(5)(int)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^(-s)`
For aceleration: `a=(dv)/(dt)=(d)/(dt)(4t-t^(2))=4-2t`.
At `t=0, a=4 ms^(-2)`
Therefore, optione (c) and (d) are correct, and options (a) and (b) and wrong.


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