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?

11 Views | Posted 3 months ago

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Answered by

Payal Gupta | Contributor-Level 10

3 months ago

This is a long answer type question as classified in NCERT Exemplar

(a) for maximum velocity dv/dt=0

d/dt (6t-2t²)=0

6-4t=0 t= 6/4=1.5s

(b) v=6t-2t²

ds/dt=6t-2t²

ds=6t-2t²dt

distance in 3s, S= $\int_{0}^{3} (6 t - {2 t}^{2}) d t = [3 t^{2} - \frac{2}{3} t^{3}]$ ³₀

s= 27-18=9m

average velocity = distance /time =9/3 = 3m/s

x= 6t-2t²

3=6t-2t²

After solving we get t= 9/4s approx.

(c) in periodic motion when velocity is zero

0=6t-2t²

0=t (6-2t)

So t=0, 3 sec

(d) distance covered from 0 to 3s=9m

distance covered in 3 to 6s= $\int_{3}^{6} (18 - 9 t + t^{2}) d t$

S= (18t- $\frac{9 t^{2}}{2} + \frac{t^{3}}{3}$ )⁶

S= 108-9 (18)+ $\frac{6^{3}}{3} - 18 (3) + \frac{9}{2} (9) - \frac{27}{3}$

S= -4.5m

So total distance covered = 9+ (-4.5)=4.5m

No of cycles covered in that distance =20/4.5=4.44approx

This is a long answer type question as classified in NCERT Exemplar (a)  for maximum velocity dv/dt=0d/dt (6t-2t2)=06-4t=0 t= 6/4=1.5s (b) v=6t-2t2ds/dt=6t-2t2ds=6t-2t2dtdistance in 3s, S= <math> <msubsup> <mo>∫</mo> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msubsup> <mrow> <mrow> <mfenced separators="|"> <mrow> <mrow> <mn>6</mn> <mi>t</mi> <mo>-</mo> <msup> <mrow> <mrow> <mn>2</mn> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> <mi>d</mi> <mi>t</mi> <mo>=</mo> <mo> [</mo> <mn>3</mn> <msup> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>-</mo> <mfrac> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mfrac> <msup> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> <mo>]</mo> </mrow> </mrow> </math> 30s= 27-18=9maverage velocity = distance /time =9/3 = 3m/sx= 6t-2t23=6t-2t2After solving we get t= 9/4s approx. (c) in periodic motion when velocity is zero0=6t-2t20=t (6-2t)So t=0, 3 sec (d) distance covered from 0 to 3s=9mdistance covered in 3 to 6s= <math> <msubsup> <mo>∫</mo> <mrow> <mn>3</mn> </mrow> <mrow> <mn>6</mn> </mrow> </msubsup> <mrow> <mrow> <mfenced separators="|"> <mrow> <mrow> <mn>18</mn> <mo>-</mo> <mn>9</mn> <mi>t</mi> <mo>+</mo> <msup> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> <mi>d</mi> <mi>t</mi> </mrow> </mrow> </math> S= (18t- <math> <mfrac> <mrow> <mrow> <mn>9</mn> <msup> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>+</mo> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mfrac> </math> )6S= 108-9 (18)+ <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mfrac> <mo>-</mo> <mn>18</mn> <mfenced separators="|"> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mfenced> <mo>+</mo> <mfrac> <mrow> <mrow> <mn>9</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mfenced separators="|"> <mrow> <mrow> <mn>9</mn> </mrow> </mrow> </mfenced> <mo>-</mo> <mfrac> <mrow> <mrow> <mn>27</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mfrac> </math> S= -4.5mSo total distance covered = 9+ (-4.5)=4.5mNo of cycles covered in that distance =20/4.5=4.44approx

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