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?

<p><span data-teams="true">This is a long answer type question as classified in NCERT Exemplar</span></p><p><strong> (a)</strong> &nbsp;for maximum velocity dv/dt=0</p><p>d/dt (6t-2t²)=0</p><p>6-4t=0 t= 6/4=1.5s</p><p><strong> (b)</strong> v=6t-2t²</p><p>ds/dt=6t-2t²</p><p>ds=6t-2t²dt</p><p>distance in 3s, S=<span contenteditable="false"> <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> </span>³₀</p><p>s= 27-18=9m</p><p>average velocity = distance /time =9/3 = 3m/s</p><p>x= 6t-2t²</p><p>3=6t-2t²</p><p>After solving we get t= 9/4s approx.</p><p><strong> (c)</strong> in periodic motion when velocity is zero</p><p>0=6t-2t²</p><p>0=t (6-2t)</p><p>So t=0, 3 sec</p><p><strong> (d)</strong> distance covered from 0 to 3s=9m</p><p>distance covered in 3 to 6s=<span contenteditable="false"> <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> </span></p><p>S= (18t-<span contenteditable="false"> <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> </span>)⁶</p><p>S= 108-9 (18)+<span contenteditable="false"> <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> </span></p><p>S= -4.5m</p><p>So total distance covered = 9+ (-4.5)=4.5m</p><p>No of cycles covered in that distance =20/4.5=4.44approx</p>

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