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

Explanation

For maximum velocity, `(dv)/(dt) = 0`
Given velocity, `v = 2t (3-t) = 6t - 2t ^(2)`
(a) From maximum velocity,
`(dv)/(dt) =0`
`therefore (d)/(dt) (6t - 2t ^(2)) =0`
`therefore 6- 4t =0`
`therefore t = (6)/(4) = 3/2 s = 1.5 s`
(b) From equ. (i) `v = 6t - 2t ^(2)`
`therefore (dx)/(dt) = 6t - 2t ^(2)`
`therefore (dx)/(dt) = 6t - 2t ^(2)`
`therefore dx = (6t - 2t ^(2)) dt` where x is displacement.
`therefore` Distance travelled in time interval 0 to 3 s.
`x _(1) = int _(0) ^(3) ( 6t - 2t ^(2)) dt`
`= [ ( 6t ^(2))/( 2 ) - ( 2t ^(3))/( 3) ] _(0) ^(3) = [ 3t ^(2) - (2)/(3) t ^(3) ] _(0) ^(3)`
`= 3 xx 9 - 2/3 xx 3 xx 3 xx 3=27-18 =9m`
Average velocity `= ("Displacement")/("Time")=9/3=3m//s`
Given, `x = 6t - 2t ^(2)`
`3 = 6t - 2t ^(2)`
`therefore 2t ^(2) - 6t - 3=0`
`therefore t = (6pm sqrt (6 ^(2) - 4 xx 2 xx 3 ))/(2 xx 2 ) = ( 6pm sqrt ( 36 - 24))/( 4)`
`= (6pm sqrt12)/(4) = (3pm 2 sqrt3)/(2)`
Considering positive sign only.
`therefore t = (3 + 2 sqrt3)/( 2) = (3 + 2 xx 1.732)/(2) = 9/4s`
(c) In a periodic motion when velocity is zero acceleration will be maximum putting v =0 in equ. (i),
`0= 6t - 2t ^(2)`
` therefore 0= t ( 6- 2t)`
` = t xx 2 (3-t) =0`
`therefore t = 0 or t =3 s`
(d) Distance covered in 0 to `3s = 9m`
Distance covered in
3 to `6s int _(3) ^(5) (18-9t +t ^(2)) dt`
`x _(2) = (18 t - (9t^(2))/( 2) + (t ^(3))/( 2)) _(3) ^(6)`
`= 18 xx 6- 9/2xx 6^(2) + (6 ^(3))/(3)`
`- (18 xx 3 - (9 xx 3 ^(2))/(2) + ( 3 ^(3))/( 3))` `= 108 -9 xx 18 + (6 ^(3))/( 3) `
`- 18 xx 3 + 9/2 xx 9- (27)/(3) = 108 - 18 xx 9+ (216)/(3) `
`- 54 + 4.5 xx 9-9 =- 4.5 m`
`therefore` Total distance travelled in oen cycle,
`= x _(1) + x _(2) = 9 -4.5 = 4.5 m`
Number of cycles covered,
`= ("total distance ")/("distance covered in each cycle")= (20)/(4.5) ~~4.44 ~~5`