Question

A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by v(t) = 2t (3 – t); 0 < t < 3 and v(t) = – (t – 3)(6 – t) for 3 < t < 6 s in m/s. It repeats this cycle till it reaches the height of 20 m.

1. At what time is its velocity maximum?
2. At what time is its average velocity maximum?
3. At what time is its acceleration maximum in magnitude?
4. How many cycles (counting fractions) are required to reach the top?

Answer 1

In this problem to calculate maximum velocity, we will use `(dv)/(dt)` = 0 and then the time corresponding to maximum velocity will be obtained.

Given velocity v(t) = 2t (3 – t) = 6t – 2t²  ......(i)

a. For maximum velocity `(dv(t))/(dt)` = 0

⇒ `d/(dt) (6t - 2t^2)` = 0 

⇒ `6 - 4t` = 0

⇒ t = `6/4 = 3/2`s = 1.5 s

b. From equation (i) v = 6t – 2t²

⇒ `(ds)/(dt) = 6t - 2t^2`

⇒ `ds = (6t - 2t^2)dt`

Where s is displacement

∴ Distance travelled in time intervals 0 to 3s.

s = `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

= 9 m.

Average velocity = `"Distance travelled"/"Time"`

= `9/3` = 3 m/s

Given, x = 6t – 2t²

⇒ 3 = 6t – 2t²

⇒ 2t² – 6t – 3 = 0

⇒ 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`

Considering positive sign only

t = `(3 + 2sqrt(3))/2`

= `(3 + 2 xx 1.732)/2`

= `9/4` s

c. In a periodic motion when velocity is zero acceleration will be maximum putting v = 0 in equation (i)

0 = 6t – 2t² 

⇒ 0 = t(6t – 2t)

= t × 2(3 – t) = 0

⇒ t = 0 or 3 s

d. Distance covered in 0 to 3 s = 9 m

Distance covered in 3 to 6 s = `int_3^6 (18 - 9t + t^2)dt`

= `(18 t - (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 - 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

∴ Total distance travelled in one cycle = s₁ + s₂ = 9 – 4.5 = 4.5 m

The number of cycles covered in total distance to be covered = `20/4.5` ≈ 4.44 ≈ 5.