Question

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 `10 ms^(-1)`
C. average velocity is `5//3 ms^(-1)`
D. acceleration is `4 ms^(-2)` at `t = 0`

Answer 1

Correct Answer - C::D
Average velocity `vec(v) = (underset(0)overset(5)int v dt)/(underset(0)overset(5) dt) = (underset(0)overset(5)int (4t - t^(2)) st)/(underset(0)overset(5)int dt)`
`= ([2t^(2) - (t^(3))/(3)]_(0))/(5) = (50 - (125)/(3))/(5) = (25)/(3 xx 5) = (5)/(3)`
For average speed, let us put `v = 0`, which gives `t = 0` and `t = 4s`
`:.` average speed `= (|underset(0)overset(4)int v dt| + |underset(4)overset(5)int v dt|)/(underset(0)overset(5)int dt)`
`= (|underset(0)overset(4)int (4 t - t^(2)) dt| + |underset(4)overset(5)int v dt|)/(5)`
`= ([ 2 t^(2) - (t^(3))/(3)]_(0)^(4) + [2t^(2) - (t^(3))/(3)]_(4)^(5))/(5)`
`= (|[2 t^(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) (4 t - t^(2)) = 4 - 2t`
At `t = 0, a = 4 ms^(-2)`