Ncert Exercises & Solutions

A monkey climbs up a slippery pole for 3 second and subsequently slips for 3 second.

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 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?

\begin{matrix} (a) Hint : For maximum velocity, \frac{dv (t)}{dt} = 0 . Step 1 : Calculate time at which velocity is maximum \frac{dv (t)}{dt} = 0 H e r e v (t) = 6 t - 2 t^{2} \Rightarrow \frac{d}{dt} (6 t - 2 t^{2}) = 0 \Rightarrow 6 - 4 t = 0 \Rightarrow t = \frac{6}{4} = \frac{3}{2} s = 1 .5 sec \end{matrix}

$\begin{array}{l} (a) Hint : For maximum velocity, \frac{dv (t)}{dt} = 0 . \\ Step 1 : Calculate time at which velocity is maximum \\ \frac{dv (t)}{dt} = 0 \\ H e r e v (t) = 6 t - 2 t^{2} \\ \Rightarrow \frac{d}{dt} (6 t - 2 t^{2}) = 0 \\ \Rightarrow 6 - 4 t = 0 \\ \Rightarrow t = \frac{6}{4} = \frac{3}{2} s = 1 .5 \sec \end{array}$

(b) For maximum average velocity net displacement should be maximum.

Step 2: Calculate net displacement in 0 < t < 3 sec.

$A s \frac{ds}{dt} = v (t)$ $A s \frac{ds}{dt} = v (t)$
$\Rightarrow ds = v (t) dt$ $\Rightarrow ds = v (t) dt$
$\Rightarrow s = \int_{0}^{3} (6 t - 2 t^{2}) d t$ $\Rightarrow s = \int_{0}^{3} (6 t - 2 t^{2}) d t$
$= {[\frac{6 t^{2}}{2} - \frac{2 t^{3}}{3}]}_{0}^{3} = {[3 t^{2} - t^{3}]}_{0}^{3}$ $= {[\frac{6 t^{2}}{2} - \frac{2 t^{3}}{3}]}_{0}^{3} = {[3 t^{2} - t^{3}]}_{0}^{3}$
$= 3 \times 9 - \frac{2}{3} \times 3 \times 3 \times 3$ $= 3 \times 9 - \frac{2}{3} \times 3 \times 3 \times 3$
$= 27 - 18 = 9 m$ $= 27 - 18 = 9 m$

where, s is displacement

$\begin{matrix} A v e r a g e v e l o c i t y = \frac{N e t d i s p l a c e m e n t}{T o t a l t i m e t a k e n} = \frac{9}{3} = 3 m / s . \end{matrix}$ $\begin{array}{l} A v e r a g e v e l o c i t y = \frac{N e t d i s p l a c e m e n t}{T o t a l t i m e t a k e n} \\ = \frac{9}{3} = 3 m / s . \end{array}$

Step 3: Calculate the time at which average velocity is maximum

Given,

$\begin{array}{l} v (t) = 6 t - 2 t^{2} \\ \Rightarrow 3 = 6 t - 2 t^{2} \\ \Rightarrow 2 t^{2} - 6 t - 3 = 0 \\ \Rightarrow t = \frac{6 \pm \sqrt{6^{2} - 4 \times 2 \times 3}}{2 \times 2} = \frac{6 \pm \sqrt{36 - 24}}{4} \\ = \frac{6 \pm \sqrt{12}}{4} = \frac{3 \pm 2 \sqrt{3}}{2} \end{array}$

Considering positive, sign only

$t = \frac{3 + 2 \sqrt{3}}{2} = \frac{3 + 2 \times 1 .732}{2} = \frac{9}{4} s$

(c) In a periodic motion when velocity is zero acceleration will be maximum putting v = 0. $\begin{array}{l} 0 = 6 t - 2 t^{2} \\ \Rightarrow = t (6 - 2 t) \\ = t \times 2 (3 - t) = 0 \\ \Rightarrow = 0 or 3 s \end{array}$

(d) Number of cycles = $\frac{Height of pole}{Net upward displacement per cycle} .$

Step 4: Find the distance covered in 0 to 3s

As calculated above distance covered in 0 to 3s = 9 m

Step 5: Find the distance covered in 3 to 6 sec.

$\begin{array}{l} Distance covered in 3 to 6 s = \int_{3}^{6} (18 - 9 t + t^{2}) dt \\ = {[18 t - \frac{9 t^{2}}{2} + \frac{t^{3}}{3}]}_{3}^{6} \\ = 18 \times 6 - \frac{9}{2} \times 6^{2} + \frac{6^{3}}{3} - [18 \times 3 - \frac{9 \times 3^{2}}{2} + \frac{3^{3}}{3}] \\ = 108 - 9 \times 18 + \frac{6^{3}}{3} - 18 \times 3 + \frac{9}{2} \times 9 - \frac{27}{3} \\ = 108 - 18 \times 9 + \frac{216}{3} - 54 + 4 .5 \times 9 - 9 = - 4 .5 m \end{array}$

Step 6: Find net upward displacement in one cycle

$∴$ Net upward displacement traveled in one cycle $= s_{1} + s_{2} = 9 - 4.5 = 4.5 m$

Step 7: Calculate the number of cycles required to reach the top

The number of cycles $= \frac{20}{4.5} = 4.44 = 5 .$

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