A monkey climbs up a slippery pole for ` 3 seconds` and subsequently slips for `3 seconda`. Its velocity at time (t) is given by ` v (t) =2 t (3-t) , 0 lt tgt 3 s and v (t) =- (t-30 (6-t) for 3 lt tlt 6 s in m//s`. It repeats thei cycle till it reaches the height of ` 20 s.`
(a) AT wht time is its v elocity maximum ? (b) At what time is its average velocity maximum ? (c ) At what time is its accelration maximum in magnitude ? (d) How many cycles (counting fractions ) are required to reach the top ?

Explanation

It this problem to calculate maximum velocity we will use `(dv)/(dt) =0`, then the time corresponding to maximum velocity will be obntained.
Given velocity
`v(t) = 2t (3-t) = 6t - 2t^(2)"…….."(i)`
(a) For maximum velocity `(dv(t))/(dt) = 0`
`rArr d/(dt) (6t - 2t^(2)) = 0`
`rArr 6 - 4t =0`
`rArr t = (6)/(4) = (3)/(2)s = 15s`
(b) From Eq. (i) `v = 6t - 2t^(2)`
`rArr (ds)/(dt) = 6t - 2t^(2)`
`rArr ds = (6t - 2t^(2))dt`
where,s is displacement `:.` Distane travelled in time interval 0 to 3s.
`s = int_(0)^(3) (6t - 2t^(2)) dt`
`= [(6t^(2))/(2) - (2t^(3))/(3)]_(0)^(3) = [3t^(3) - (2)/(3)t^(3)]_(0)^(3)`
`= 3 xx 9 - (2)/(3) xx 3 xx 3 xx 3`
`= 27 - 18 = 9 m`
Average velocity
` = ("Distance travelled")/("Time")`
`= 9/3 = 3 m//s`
Given, `x = 6t - 2t^(2)`
`rArr 3 = 6t - 2t^(2)`
`rArr 2t^(2) - 6t - 3 = 0`
`rArr t = (6 +- sqrt(6^(2) - 4 xx 2 xx 3))/(2 xx 2) = (6 +- sqrt(36-24))/(4)`
`= (6+- sqrt(12))/(4) = (3+- 2sqrt(3))/(2)`
(c) In a periodic motion when velocinty is zero accleration will be maximum putting `v = 0` in Eq. (i) `0 = 6t - 2t^(2)`
`rArr 0 = t (6-2t)`
`= t xx 2 (3-t) = 0`
`rArr t = 0` or `3s`
(d) Distance covered in `0` to `3s = 9m`
Distance covererd in 3 to 6s = `int_(3)^(6) (18 - 9t +t^(2)) dt`
`= (18t - (9t^(2))/(2) + (t^(3))/(3))_(3)^(6)`
`= 18 xx 6 - (9)/(2) xx 6^(2) + (6^(3))/(3) - (18 xx 3 - (9 xx 3^(2))/(2) + (3^(3))/(3))`
`= 108-9xx18+(6^(3))/(3) - 18 xx 3 + (9)/(2) xx 9 - (27)/(3)`
`= 108-18 xx 9 + (216)/(3) - 54+45xx9-9 = - 45 m`
`:.` Total distance travelled in one cycle `= s_(1) + s_(2) = 9 - 45 = 45m`
Number of cycles covered in total distance to be covered `= (20)/(4.5) = 4.44 ~~ 5`