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

` v (t) = 2 t (3- t) 6 t-2 t^2`
` a= (dv (t))/(dt) = 6- 4 t`
(a) When velocity v (t) maximum, ` (du (t))/(dt) =0 so 6- 4 t=0 or t= (3 //2)s`
(b) Fro, (i), (ds)/(dt) = 6 t -2 t^2 or ds =( 6 t- 2 t^2) at`
Destance travelled in time interval ` 0 to 3 seconds` will be
S_1 = int ( 6 t- 2t^2) dt = ( (6t^2)/2 - (2t^3)/3)_0^3 = (( 6 xx3^(2) )/2 - (2 xx 3^3)/3 ) =9 m`
v_(av) = s_1 /t = 9/3 = 3 m//s
From (i), ` 3 = 6 t- 2 t^2 or 2 t^2-6 t+ 3 = 0`.
:. ` t= ( 6 +- sqrt(36 - 24)/4 = (6 +- sqrt12)/4 = (9.464)/4 ~~ 9/4 s`
(c ) In `S_1 = int _0^3 (6t- 2 t^2) dt = ((6t^2)/2 - (2t^3)/3 )_0^3 = 3 xx 3 ^2 - 2/3 xx 3^3 = 27 - 18 =9 m`
`S_1 = int _3 &6 - (t-3) (6-) = int_3^6 (18 t - 99t^2)/2 + t^3 /3 )_3^6`
`= (18 xx 6- (9 xx 6^3) /2 = 6^3/3 + 6^3/3) - ( 18 xx 3 - (9 xx 3^2 ) /2 + 3^3/3)`
` =18 - 22 .5 =- 4.5 m`
Total distance travelled in one cycle `S_1 = S_2 = 9 - 4. 5 = 4.5 m`.
No. of cycles covered in total distance to be coverd ` = 920)/(4.5) ~~ 4.44 =5`.