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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.

(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?

1 Answer

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Given, v(t) = 2t(3 – t) = 6t – 2t

(a) For maximum velocity,

\(\frac{dv(t)}{dt}\) = 0

\(\frac{d}{dt}\) (6t − 2t2 ) = 0

⇒ 6 − 4t = 0⇒ t =\(\frac{6}{4}\) = \(\frac{3}{2}\)s = 1.5 s

(b) \(\frac{ds}{dt}\) = 6t − 2t2 [s = displacement]

ds = (6t – 2t2 ) dt

∴ Distance travelled in time interval to 0 to 3s.

s = \(\int_0^3(6t -2t^2)dt\) 

= \({[\frac{6t^2}{2}-\frac{2t^3}{3}]}^3_0={[3t^2-\frac{2}{3}t^3]}^3_0\)  

= 3×9− \(\frac{2}{3}\) × 3 × 3 × 3 = 27 – 18 = 9 m.

Average velocity = \(\frac{9}{3}\) = 3 m/s

As, vavg = 6t − 2t2 ⇒ 3 = 6t – 2t2⇒2t2 − 6t − 3 = 0

⇒  \(t=\frac{6\pm\sqrt{6^2-4\times2\times3}}{2\times2}=\frac{6\pm\sqrt{36-24}}{4}=\frac{6\pm\sqrt {12}}{4}=\frac{3\pm\sqrt 3}{2}\) 

Taking ‘+’ ve sign we get \(t_1=\frac{3+\sqrt 3}{2}\) = 2.36 sec.

Taking ‘−‘ ve sign we get \(t^2=\frac{3-\sqrt 3}{2}\) = 0.633 sec

which is less than the least of the clock so it cannot be measured,

∴ rejecting t2, the average velocity is maximum at 2.36 sec.

(c) Acceleration will be maximum when velocity is zero

∴ (6t − t2 ) = 0

t(6−2t) = 0

t = 0, 3 sec

(d) \(s_1=\int_0^3(6t - 2t^2)dt\)

= \(\left.\frac{6t^2}{2}-\frac{2t^3}{3}\right|_0^3\)

= 9 m

\(s_2=\int_3^6(-(t-3)(6-t))dt\)

= \(\int_3^6-(6t-t^2-18+3t)dt\)

= \(\int_3^6-(t^2+9t-18)dt\)

= \(\left.\frac{t^3}{3}-\frac{9t^2}{2}+18t\right|_3^6\)

= −4.5 m

Total distance travelled in one cycle is

= 9 – 4.5 m = 4.5 m

No. of cycles = \(\frac{20}{4.5}\) = 4.44 cycles ≈ 5 cycles

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