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