If velocity of a partical moving along a straight line changes sinusoidally with time as shown in given graph. Find the average speed over time interval `t = 0` to `t= 2 (2 n - 1)` second, `n` being any positive interget.

Answer

`(8)/(pi)m//s`

Explanation

Method `- 1`
Average speed `= ("total distance travelled")/("total time taken") = (S)/(2(2n - 1))`
Here `Delta t = 2 (2n - 1) = 4n - 2 = 4(n - 1) + 2`
From the graph it is clear that the Time period `T = 4` sec.
`:. Deltat = (n - 1)T + (T)/(2)`
Total distance travelled in one time period is `= 4A`
where `'A'` is amplitude
`:.` Total distance travelled in `Deltat` is
`S = (n - 1)4A + 2A = (2n - 1)2A`
`ltvgt = ((2n - 1)2A)/(2(2n - 1)) = A` and `omegaA = v_(max) = 4 rArr (2pi)/(4) = 4rArr A = (8)/(pi) , ltvgt = (8)/(pi)m//s`
Method `- 2`
If can be observed from the graph, that average speed in time interval `t = 0` to `t = 2` sec is same as that in intervals `t = 0` to `t = 4` sec., `t = 0` to `t = 8` sec., `t = 0` to `t = 12` sec.......
ot `t = 0` to `t = 2(2n - 1)` seconds
The speed as funcation of time is
`v = |4 sin ((2pi)/(T))t| = |4sin ((2pi)/(4))t| = |4 sin ((pit)/(2))|`.
`The average speed in time interval ltbr. `t = 0` to `t = 2` sec is
`overset(bar)(v)= (overset(2)underset(0)(int)vdt)/(overset(2)underset(0)(int)dt) = (overset(2)underset(0)(int)4sin'(pit)/(2)dt)/(2) = (8)/(pi)m//s`