Physics Ncert Exemplar 2019 Solutions for Class 11 Science Physics Chapter 3 - Motion In A Straight Line

Answer:

Average velocity is zero when total displacement is zero. In graph shown in option (b) at time t = 0  and at time t = T (suitably chosen)  position is same so displacement is zero.


So average velocity from t = 0 to t = T is zero.
Hence, the correct answer is option (b). 

Answer:

Lift is moving downwards so x is negative i.e x < 0.
Here we can eliminate other options easily. Moreover velocity is downwards so v is also negative i.e v < 0.

Hence the correct answer is option (a)

Answer:

Here instantaneous speed can vary from zero to v₀. So maximum distance v₀T can be travelled in either positive or negative direction. Hence the displacement x in time T satisfies $-$v₀T < x < v₀T.
Hence the correct answer is option (b).

Answer:

Let time taken to travel first half and the second half distance of the journey be t₁ and t₂ respectively.

$$\begin{gathered} \mathrm{Time} \mathrm{taken} \mathrm{to} \mathrm{travel} \mathrm{first} \mathrm{half} \mathrm{distance}, t_1 = \frac{L/2}{v_1} = \frac{L}{2v_1} \\ \mathrm{Time} \mathrm{taken} \mathrm{to} \mathrm{travel} \mathrm{second} \mathrm{half} \mathrm{distance}, t_2 = \frac{L/2}{v_2} = \frac{L}{2v_2} \\ \mathrm{Total} \mathrm{time} = \frac{L}{2v_1} + \frac{L}{2v_2} \\ \mathrm{Since} \mathrm{average} \mathrm{speed} = \frac{\mathrm{Total} \mathrm{distance}}{\mathrm{total} \mathrm{time}} = \frac{L}{\frac{L}{2v_1} + \frac{L}{2v_2}} =\frac{2v_1{v}_2}{v_1+v_2} \end{gathered}$$
Hence, the correct answer is option (c).

Answer:

$$\begin{gathered} \mathrm{Since} x={\left(t-2\right)}^2 \\ \mathrm{then} \mathrm{velocity}, v=\frac{dx}{dt} = 2\left(t-2\right) \\ \mathrm{and} \mathrm{at} t=0, v=-4 \mathrm{m}/\mathrm{s}, \\ v=0 at t=2 \mathrm{s} \\ v=4 \mathrm{m}/\mathrm{s} \mathrm{at} t=4 \mathrm{s} \\ \mathrm{Also}, \\ \mathrm{Acceleration} = \frac{dv}{dt}= 2 \mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Now} \mathrm{v}-\mathrm{t} \mathrm{graph} \mathrm{of} \mathrm{the} \mathrm{particle} \mathrm{is} \mathrm{shown} \mathrm{in} \mathrm{the} \mathrm{figure}. \\ \mathrm{Distance} \mathrm{travelled} = \mathrm{area} \mathrm{OAD} + \mathrm{area} \mathrm{ABC} \\ = \frac{1}{2}\times 4\times 2 + \frac{1}{2}\times 4\times 2 = 8 \mathrm{m}. \end{gathered}$$



Hence the correct answer is option (b).

Answer:

Let the total distance covered be L.
$$\begin{gathered} \mathrm{So}, \mathrm{velocity} \mathrm{of} \mathrm{girl} = \frac{L}{t_1} \\ \mathrm{velocity} \mathrm{of} \mathrm{escalator} = \frac{L}{t_2} \\ \mathrm{Effective} \mathrm{velocity} \mathrm{of} \mathrm{girl} \mathrm{on} \mathrm{escalator} = \frac{L}{t_1} + \frac{L}{t_2} \\ \mathrm{if} \mathrm{t} \mathrm{is} \mathrm{the} \mathrm{time} \mathrm{taken}, \mathrm{then} \\ \frac{L}{t}=\frac{L}{t_1} + \frac{L}{t_2} \\ \Rightarrow t=\frac{{\mathrm{t}}_1{\mathrm{t}}_2}{{\mathrm{t}}_1+{\mathrm{t}}_2} \end{gathered}$$
Hence the correct answer is option (c).

Answer:

Quantity B may represent time. Thus option (a) is true.
If motion is uniform then velocity is constant. Thus option (b) is false.
If motion is uniform then displacement-time graph is straight line. Thus option (c) is true.
If motion is accelerated then velocity-time graph is a straight line with positive slope. Thus option (d) is true.

Hence the correct answer is option (a), (c) and (d)

Answer:

At point A the graph is parallel to time axis, hence v = $\frac{dx}{dt}$= 0. Hence we can say that the particle is starting from rest. Thus option (a) is correct.
At B slope is zero i.e., velocity is zero but particle changes direction i.e., acceleration is opposite to velocity i.e., negative. Thus option (b) is false.
At C, the graph changes slope, hence velocity also changes. As graph at C is almost parallel to time axis so velocity is zero and acceleration is also zero.
Since displacement is negative from A to D, average velocity is negative from A to D.  Thus option (d) is false.
Magnitude of slope at E is less than magnitude of slope at D.  Thus option (e) is true. 

Hence the correct options are (a), (c) and (e).

Answer:

Given equation,
$$\begin{gathered} x\mathit{ }\mathit{=}\mathit{ }t\mathit{-}\sin t \\ \mathrm{So}, \mathrm{velocity}, v\mathit{=}\frac{\mathit{d}x}{\mathit{d}t}\mathit{=}\mathit{1}\mathit{-}\cos t \\ \mathrm{acceleration}, \mathrm{a}=\frac{dv}{dt}= \sin t \\ \mathrm{so}, x>0 \mathrm{for} \mathrm{all} \mathrm{values} \mathrm{of} t>0. \\ v \mathrm{can} \mathrm{be} \mathrm{zero} \mathrm{if} \cos t=1. \\ a \mathrm{can} \mathrm{be} \mathrm{zero} \mathrm{if} \sin t=0. \\ \mathrm{if} t=0, v\left(0\right) = 1-1=0 \\ \mathrm{if} t=\mathrm{\pi }, v\left(\mathrm{\pi }\right)=1-\cos \left(\mathrm{\pi }\right)= 2 \end{gathered}$$

Hence the correct options are (a) and (d) only.

Answer:

Spring force = −kx where x is the displacement from natural length of spring. So the force is maximum when x is maximum. Hence acceleration is maximum when block is just released. Hence option (a) is correct.
Total Energy is conserved i.e KE+PE=TE.
At equilibrium position, KE is max since PE = $\frac{1}{2}k{x}^2$ and x = 0. So speed is maximum at equilibrium position. Hence option (c) is correct.
If the fixed end of the spring is taken as origin than speed is zero when the displacement is maximum or minimum. Hence option (d) is false if fixed end of spring is taken as origin.

Hence the correct options are (a) and (c).

Answer:

Let v_b and v_t be the speed of ball and train with respect to ground.
Then speed of ball with respect to train is: $$\begin{gathered} v_{bt}=v_b-v_t \\ \Rightarrow v_b=v_{bt}+v_t \\ \mathrm{Now} \mathrm{before} \mathrm{collision}, v_b = 10+1 = +11 \mathrm{m}/\mathrm{s} \\ \mathrm{after} \mathrm{colision} ,v_b = -1+10 = +9 \mathrm{m}/\mathrm{s} \\ \mathrm{In} \mathrm{both} \mathrm{the} \mathrm{cases} \mathrm{speed} \mathrm{of} \mathrm{ball} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{ground} \mathrm{is} \mathrm{positive} \mathrm{i}.\mathrm{e} \mathrm{to} \mathrm{the} \mathrm{ground} \mathrm{observer}, \mathrm{the} \mathrm{ball} \mathrm{will} \mathrm{never} \mathrm{change} \mathrm{is} \mathrm{direction} \mathrm{and} \mathrm{will} \mathrm{always} \mathrm{appear} \mathrm{to} \mathrm{move} \mathrm{in} \mathrm{same} \mathrm{direction} \mathrm{with} \mathrm{two} \mathrm{different} \mathrm{speeds} \mathrm{after} \mathrm{every} 10 \mathrm{seconds}. \\ \Rightarrow \mathrm{option} \left(\mathrm{a}\right) \mathrm{is} \mathrm{false}. \\ \mathrm{Speed} \mathrm{of} \mathrm{the} \mathrm{ball} \mathrm{will} \mathrm{change} \mathrm{after} \mathrm{every} 10 \mathrm{s}. \\ \Rightarrow \mathrm{option} \left(\mathrm{b}\right) \mathrm{is} \mathrm{true}. \\ \mathrm{Since} \mathrm{ball} \mathrm{moves} \mathrm{at} 11 \mathrm{m}/\mathrm{s} \mathrm{for} 10 \mathrm{s} \mathrm{and} \mathrm{then} 9 \mathrm{m}/\mathrm{s} \mathrm{for} \mathrm{next} 10 \mathrm{s}. \\ \mathrm{Average} \mathrm{speed} \mathrm{of} \mathrm{ball} = \frac{\mathrm{total} \mathrm{distance}}{\mathrm{total} \mathrm{time}} =\frac{11\times 10+9\times 10}{10+10} =\frac{200}{20}=10 \mathrm{m}/\mathrm{s} \\ \Rightarrow \mathrm{option} \left(\mathrm{c}\right) \mathrm{is} \mathrm{true}. \\ \mathrm{Since} \mathrm{speed} \mathrm{of} \mathrm{train}, \mathrm{ball} \mathrm{and} \mathrm{observer} \mathrm{are} \mathrm{constant}, \mathrm{their} \mathrm{accelerations} \mathrm{are} \mathrm{zero}. \\ \Rightarrow \mathrm{option} \left(\mathrm{d}\right) \mathrm{is} \mathrm{true}. \end{gathered}$$

Hence the correct options are (b), (c) and (d).

Answer:

For graph (a), at point P, x = 0 at time t > 0. so displacement is zero at one point for t > 0. 

So (a) matches with (iii).
For graph (b), $\frac{dx}{dt}$ is zero at point Q. So v = 0 at only one point. Acceleration i.e., $\frac{d^{2}x}{d{t}^2}$ = 0 at point R. So, a = 0 at only one point.
Hence, (b) matches with (ii).

For graph (c) $\frac{dx}{dt}$ < 0 and $\frac{d^{2}x}{d{t}^2}$ > 0. 
So (c) matches with (iv).

For graph (d) $\frac{dx}{dt}$ > 0 and $\frac{d^{2}x}{d{t}^2}$< 0. 
so (d) matches with (i).





Hence the correct matching sequence is (a) (iii), (b) (ii), (c) iv, (d) (i).
 

Answer:

As the ball collide with bat, it will get deformed.



Impulsive forces will act on the ball. Impulsive forces will be restoring forces due to deformation.
Magnitude of restoring forces will be directly proportional to the deformation.
Hence force will be more when the deformation will be more.
So force will increase for half the duration of collision and force will decrease for next half.
Since $a=\frac{F}{m}$, the graph will look like:

Answer:

Answer:

Let us choose an equation,
$$\begin{gathered} x\left(t\right)=A+ B e^{-kt} \mathrm{where} A > B, k > 0 \mathrm{are} \mathrm{suitably} \mathrm{chosen} \mathrm{positive} \mathrm{constants}. \\ v=\frac{dx}{dt}=-Bk e^{-kt} (\mathrm{Which} \mathrm{is} \mathrm{negative}) \\ a=\frac{dv}{dt}=B{k}^2 e^{-kt} (\mathrm{Which} \mathrm{is} \mathrm{positive}) \end{gathered}$$

Answer:

If speed is constant,
Then acceleration, a = 0.
i.e g − bv = 0
or, v = $\frac{g}{b}$
Hence, the value of constant speed is $\frac{g}{b}$.

Answer:

At point A, velocity is zero. At point B velocity, v = −gt.

At point C, velocity is −gt just before collision and smaller than +gt just after collision. 

At point E velocity becomes zero and motion continues.

The ball is released and is falling under gravity.

Acceleration is –g , except for the short time intervals in which the ball collides with ground, and when the impulsive force acts and produces a large acceleration.

Answer:

(a). Displacement,
$$\begin{gathered} x\left(t\right)=x_0 (1 – e^{–\gamma t} ) \\ \mathrm{at} t=0, x=x_0(1-1) =0 \\ x \mathrm{is} \mathrm{maximum} \mathrm{when} t=\infty . \\ {x}_{max}=x_0 \\ x \mathrm{is} \mathrm{minimum} \mathrm{when} t\mathit{=}\mathit{0}\mathit{,}\mathit{ }{x}_{min}\mathit{=}\mathit{ }\mathit{0} \\ \mathrm{Similarly}, \mathrm{velocity} =\frac{dx}{dt}=x_0\gamma e^{–\gamma t} \\ \mathrm{at}, t=0, v=x_0\gamma \end{gathered}$$
So, the particle stats from position x = 0 with velocity $x_o\gamma$.

(b).
$$\begin{gathered} v \mathrm{is} \mathrm{maximum} \mathrm{when} t=0. \\ {v}_{max}\mathit{=}{x}_{\mathit{0}}\gamma \\ v \mathrm{is} \mathrm{minimum} \mathrm{when} t\mathit{=}\mathit{\infty }\mathit{.}\mathit{ } \\ {v}_{min}\mathit{=}\mathit{0} \\ \mathrm{Now}, \mathrm{acceleration}, \\ a\mathit{=}\frac{\mathit{d}v}{\mathit{d}t}\mathit{=}\mathit{-}{x}_{\mathit{0}}{\gamma }^{\mathit{2}}{e}^{\mathit{–}\gamma t} \\ a \mathrm{is} \mathrm{maximum} \mathrm{when} t\mathit{=}\mathit{\infty }\mathit{.}\mathit{ } \\ {a}_{max}\mathit{=}\mathit{0} \\ a \mathrm{is} \mathrm{minimum} \mathrm{when} t\mathit{=}\mathit{0}\mathit{.}\mathit{ } \\ {a}_{min}\mathit{=}\mathit{-}{x}_{\mathit{0}}{\gamma }^{\mathit{2}} \end{gathered}$$

Answer:

Relative speed of cars = 18+27 = 45 km/h,
time required to meet =  $\frac{36}{45}$ = 0.8 h 
Thus, total distance covered by the bird = 36  × 0.8  = 28.8 km

Total displacement of the bird = distance travelled by first car = 18 × 0.8 = 14.4 Km

Answer:

$$\begin{gathered} \mathrm{Suppose} \mathrm{that} \mathrm{the} \mathrm{fall} \mathrm{of} 9 \mathrm{m} \mathrm{will} \mathrm{take} \mathrm{time} \mathrm{t}. \\ \mathrm{Hence} h=u_{y}t-\frac{g{t}^2}{2} (u_y=\mathrm{component} \mathrm{of} \mathrm{initial} \mathrm{velocity} \mathrm{in} \mathrm{downward} \mathrm{direction}) \\ \mathrm{Since} u_{oy}= 0, \\ \Rightarrow t=\sqrt{\frac{2h}{\mathrm{g}}}=\sqrt{\frac{2\times 9}{10}}= 1.34 \mathrm{s}. \\ \mathrm{In} \mathrm{this} \mathrm{time}, \mathrm{the} \mathrm{distance} \mathrm{moved} \mathrm{horizontally} \mathrm{is} \\ R = u_{x}t = 9 \mathrm{m}/\mathrm{s} \times 1.34 \mathrm{s} = 12.06 \mathrm{m}. (u_x=\mathrm{component} \mathrm{of} \mathrm{initial} \mathrm{velocity} \mathrm{in} \mathrm{horizontal} \mathrm{direction}) \end{gathered}$$
Since, the R > 10 m, so the man will safely land on the next building.

Answer:

Both are free falling. Hence, there is no acceleration of one w.r.t. another. Therefore, relative speed remains constant (=40 m/s ).
 Velocity of ball dropped after time t,
v_d​ = gt (downwards)    ...(1)
Velocity of ball thrown after time t,
v_t = 40 − gt (upwards)     ...(2)
So, relative speed of ball,
​ v_dt = v_d + v_t = gt + 40 - gt
      = 40 m/s.

Answer:

$$\begin{gathered} \mathrm{Using} \mathrm{the} \mathrm{eqn} \mathrm{of} \mathrm{straight} \mathrm{line} y\mathit{=}mx+c, \mathrm{we} \mathrm{can} \mathrm{write} \\ v\mathit{-}{v}_{\mathit{o}}\mathit{=}\left(\frac{-v_o}{x_o}\right)\mathit{ }x\mathit{ } \\ v=\left(\frac{-v_o}{x_o}\right) x + v_o \\ \mathrm{Now} \mathrm{acceleration} \mathrm{is} \mathrm{given} \mathrm{by}, a =\frac{\mathit{d}v}{\mathit{d}t} \\ a=\left(\frac{-v_o}{x_o}\right)\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}+0 \\ a = {\left(\frac{-v_o}{x_o}\right)}^{2}x - \left(\frac{-{v_o}^2}{x_o}\right)\mathit{ } \\ \mathrm{The} \mathrm{variation} \mathrm{of} \mathrm{a} \mathrm{with} \mathrm{x} \mathrm{is} \mathrm{shown} \mathrm{in} \mathrm{the} \mathrm{figure}. \mathrm{It} \mathrm{is} \mathrm{a} \mathrm{straight} \mathrm{line} \mathrm{with} \mathrm{a} \mathrm{positive} \mathrm{slope} \mathrm{and} \mathrm{a} \mathrm{negative} \mathrm{intercept}. \end{gathered}$$


 

Answer:

$$\begin{gathered} \left(\mathrm{a}\right) \mathrm{Initial} \mathrm{speed} \mathrm{of} \mathrm{drop}, u=0. \\ \mathrm{So}, v^2-0^2=2gh =\sqrt{2\times 10\times 1000}=100\sqrt{2} \mathrm{m}/\mathrm{s} \approx 510 \mathrm{km}/\mathrm{h} \\ \left(\mathrm{b}\right) \mathrm{Mass} \mathrm{of} \mathrm{drop} = \mathrm{volume}\times \mathrm{density} = \frac{4}{3}\pi r^3\times 1000 =\frac{4}{3}\times \frac{22}{7}\times \left(\frac{4}{2}\times {10}^{-3}\right)\times {10}^3 \\ \approx 3.4\times {10}^{-5} \mathrm{kg}. \\ \mathrm{Momentum} = 3.4\times {10}^{-5}\times 510 \approx 5\times {10}^{-3} \mathrm{kgm}/\mathrm{s} \\ \left(\mathrm{c}\right) \mathrm{Time} \mathrm{required} \mathrm{to} \mathrm{flatten} \mathrm{the} \mathrm{drop} = \mathrm{Time} \mathrm{taken} \mathrm{by} \mathrm{the} \mathrm{drop} \mathrm{to} \mathrm{travel} \\ \mathrm{the} \mathrm{distance} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{diameter} \mathrm{of} \mathrm{the} \mathrm{drop} \mathrm{near} \mathrm{the} \mathrm{ground}. \\ \mathrm{Time}, t\mathit{=}\mathit{ }\frac{D}{v}=\frac{4\times {10}^{-3}}{100\sqrt{2}}=0.028\times {10}^{-3} \mathrm{s} =2.8 \mathrm{\mu s} \approx 3 \mathrm{\mu s} \\ \left(\mathrm{d}\right) \mathrm{Force} = \frac{\mathit{∆}p}{\mathit{∆}t}=\frac{4.7\times {10}^{-3}}{2.8\times {10}^{-5}}\approx 168 \mathrm{N}. \\ \left(\mathrm{e}\right) \mathrm{Radius} \mathrm{of} \mathrm{the} \mathrm{umbrella} =\frac{1}{2} \mathrm{m}, \mathrm{area} \mathrm{of} \mathrm{umbrella} =\pi {\left(\frac{1}{2}\right)}^2\approx 0.8 {\mathrm{m}}^2 \\ \mathrm{With} \mathrm{average} \mathrm{separation} \mathrm{of} 5 \mathrm{cm}, \mathrm{number} \mathrm{of} \mathrm{drops} \mathrm{that} \mathrm{will} \mathrm{fall} \mathrm{almost} \mathrm{simultaneously} \mathrm{is} \\ = \frac{0.8}{{\left(5\times {10}^{-2}\right)}^2}\approx 320. \\ \mathrm{Net} \mathrm{force} = 320\times 168 =53760 \mathrm{N}\approx 54000 \mathrm{N}. \end{gathered}$$
 

Answer:

Speed of the motor car,

$$\begin{gathered} 72 \mathrm{km}/\mathrm{h} = 20 \mathrm{m}/\mathrm{s} \\ \mathrm{Regardation} \mathrm{of} \mathrm{truck} =\frac{20-0}{5}=4 {\mathrm{ms}}^{-2} \\ \mathrm{Required} \mathrm{regardation} \mathrm{of} \mathrm{car} =\frac{20-0}{3} {\mathrm{ms}}^{-2} \\ \mathrm{Let} \mathrm{the} \mathrm{truck} \mathrm{be} \mathrm{at} \mathrm{a} \mathrm{distance} s \mathrm{from} \mathrm{the} \mathrm{car} \mathrm{when} \mathrm{breaks} \mathrm{are} \mathrm{applied} \mathrm{and} t \mathrm{be} \mathrm{the} \mathrm{time} \\ \mathrm{taken} \mathrm{to} \mathrm{cover} \mathrm{this} \mathrm{distance}. \\ \mathrm{As} \mathrm{human} \mathrm{response} \mathrm{time} \mathrm{is} 0.5 \mathrm{s}, \mathrm{car} \mathrm{will} \mathrm{cover} \mathrm{some} \mathrm{distance} \mathrm{with} \mathrm{uniform} \mathrm{velocity}. \\ \mathrm{So} \mathrm{the} \mathrm{time} \mathrm{for} \mathrm{retarded} \mathrm{motion} \mathrm{of} \mathrm{car} \mathrm{is} (t-0.5) \mathrm{s}. \\ \mathrm{Velocity} \mathrm{of} \mathrm{car} \mathrm{after} \mathrm{time} \left(t\right)\mathit{=}u\mathit{-}at =20-\frac{20}{3}\left(t-0.5\right) \\ \mathrm{Velocity} \mathrm{of} \mathrm{truck} \mathrm{after} \mathrm{time} \left(t\right) =20-4\mathrm{t} \\ \mathrm{To} \mathrm{avoid} \mathrm{collision}, 20-\frac{20}{3}\left(t-\mathrm{0}\mathrm{.}\mathrm{5}\right)\mathit{=}20-4t\mathit{ }\mathit{\Rightarrow }t=1.25 \mathrm{s}. \\ \mathrm{Distance} \mathrm{travelled} \mathrm{by} \mathrm{truck} \mathrm{in} \mathrm{time} \left(t\right) \\ \mathit{ }\mathit{ }=\mathit{ }ut-\frac{1}{2}a{t}^2=\left(20\times 1.25\right)-\frac{1}{2}\times 4\times {\left(1.25\right)}^2=21.875 \mathrm{m}. \\ \mathrm{Distance} \mathrm{travelled} \mathrm{by} \mathrm{car} \mathrm{in} \mathrm{time}\mathit{ }\left(t\right) = d_1+d_2 \\ \mathrm{where} d_1 \mathrm{is} \mathrm{distance} \mathrm{travelled} \mathrm{by} \mathrm{car} \mathrm{without} \mathrm{retardation} \mathrm{in} 0.5 \mathrm{s}=\left(20\times 0.5\right) \mathrm{m} \\ \mathrm{and}, \mathrm{where} d_2 \mathrm{is} \mathrm{distance} \mathrm{travelled} \mathrm{by} \mathrm{car} \mathrm{without} \mathrm{retardation} \mathrm{in} 0.5 \mathrm{s}= 20\left(1.25-0.5\right)-\frac{1}{2}\left(\frac{20}{3}\right)\left(1.25-0.5\right) \\ \mathrm{So}, \mathrm{Distance} \mathrm{travelled} \mathrm{by} \mathrm{car} \mathrm{in} \mathrm{time} t=23.125 \mathrm{m} \\ \mathrm{Hence}, \mathrm{to} \mathrm{avoid} \mathrm{the} \mathrm{collision} \mathrm{with} \mathrm{the} \mathrm{truck}, \mathrm{the} \mathrm{car} \mathrm{must} \mathrm{maintain} \mathrm{a} \mathrm{distance} \mathrm{of} \left(23.125-21.875\right)=1.25 \mathrm{m} \mathrm{from} \mathrm{the} \mathrm{truck}. \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{a}\right) \mathrm{For} \mathrm{maximum} \mathrm{velocity}, \frac{dv}{dt}=0 \\ \mathrm{During} \mathrm{climbing} \mathrm{up} \mathrm{the} \mathrm{pole}, \\ \frac{d}{dt}\mathit{\left[}\mathit{2}t\mathit{(}\mathit{3}\mathit{-}\mathit{t}\mathit{)}\mathit{\right]}\mathit{=}\frac{d}{dt}\mathit{(}\mathit{6}t\mathit{-}\mathit{2}{t}^{\mathit{2}}\mathit{)}=0 \\ \mathrm{or} 6-4t=0 \mathrm{or}\mathit{ }t=\frac{3}{2} \mathrm{s}. \\ \mathrm{S}o \mathrm{during} \mathrm{climbing} \mathrm{up} \mathrm{the} \mathrm{pole}, \max \mathrm{velocity} = 6\times 1.5-2{\left(5\right)}^2=4.5 \mathrm{m}/\mathrm{s} \mathrm{at} \mathrm{time} t=1.5 \mathrm{s}. \\ \left(\mathrm{b}\right) \mathrm{Velocity} \mathrm{is} \mathrm{positive} \mathrm{for} \mathrm{first} 3 \mathrm{seconds} \mathrm{and} \mathrm{negative} \mathrm{for} t=3 \mathrm{to} t=6 \mathrm{s}. \\ \mathrm{So}, \mathrm{monkey} \mathrm{moves} \mathrm{up} \mathrm{during} \mathrm{first} 3 \mathrm{s} \mathrm{and} \mathrm{falls} \mathrm{down} \mathrm{during} \mathrm{next} 3 \mathrm{s}. \\ \mathrm{Displacement} \mathrm{of} \mathrm{monkey} \mathrm{for} \mathrm{first} 3 \mathrm{s} \mathrm{as} \mathrm{a} \mathrm{funtion} \mathrm{of} t\mathit{,} \\ \mathit{ }x\mathit{\left(}t\mathit{\right)}\mathit{ }\mathit{=}\mathit{ }\int vdt \\ \mathrm{or} \mathit{ }x\mathit{\left(}t\mathit{\right)}\mathit{ } =\int \mathit{(}\mathit{6}t\mathit{-}\mathit{2}{t}^{\mathit{2}}\mathit{)}dt =3t^2-\frac{2}{3}{t}^3 \\ \mathrm{Average} \mathrm{velocity} \mathrm{of} \mathrm{monkey} \mathrm{at} \mathrm{any} \mathrm{time} = \frac{\int vdt}{\int dt}=\frac{3t^2-\frac{2}{3}{t}^3}{t}=3t-\frac{2}{3}{t}^2 \\ \mathrm{For} \max \mathrm{avg} \mathrm{velocity}, \\ \frac{\mathrm{d}}{dt}\left(3t-\frac{2}{3}{t}^2\right)=0 \\ \mathrm{or}, t=2.25 \mathrm{s}. \\ \left(\mathrm{c}\right) \\ \mathrm{for} 0<t<3, \mathrm{a} = \frac{\mathrm{d}v}{dt} =\frac{\mathrm{d}}{\mathrm{dt}}\left(6t-2t^2\right)=6-4t. \\ \mathrm{Acceleartiion} \mathrm{is} \mathrm{maximum} \mathrm{at}\mathit{ }t=0 \mathrm{s}. \\ {a}_{max}=6 \mathrm{m}/{\mathrm{s}}^2. \\ \mathrm{for} 3<t<6, \mathrm{a} =\frac{\mathrm{d}}{dt}\left(t^2-9t+18\right) =2t-9. \\ \mathrm{Acceleration} \mathrm{is} \max \mathrm{at} t=6 \mathrm{s}. \\ a_{max}=3 \mathrm{m}/{\mathrm{s}}^2 \\ \left(\mathrm{d}\right) \\ \mathrm{Distance} \mathrm{travelled} \mathrm{in} \mathrm{initial} 3 \mathrm{seonds}, \\ s ={\int }_0^3\left(6t-2t^2\right)dt ={\left[3t^2-\frac{2}{3}{t}^3\right]}_0^3 \\ = 27-18=9 \mathrm{m} \\ \mathrm{Distance} \mathrm{travelled} \mathrm{in} \mathrm{next} 3 \mathrm{seonds}, \\ \mathrm{s} ={\int }_3^6\left({\mathrm{t}}^2-9\mathrm{t}+18\right)\mathrm{dt} =4.5 \mathrm{m} \\ \mathrm{Hence} \mathrm{the} \mathrm{monkey} \mathrm{moves} 4.5 \mathrm{m} \mathrm{up} \mathrm{in} \mathrm{first} \mathrm{cycle}. \mathrm{He} \mathrm{will} \mathrm{reach} 9 \mathrm{m} \mathrm{in} \mathrm{second} \mathrm{cycle} \mathrm{and} 13.5 \mathrm{m} \mathrm{in} \mathrm{third} \mathrm{cycle}. \\ \mathrm{Now} t\mathrm{he} \mathrm{monkey} \mathrm{only} \mathrm{needs} \mathrm{to} \mathrm{move} 6.5 \mathrm{m} \mathrm{to} \mathrm{reach} \mathrm{top}. \mathrm{so} \mathrm{he} \mathrm{will} \mathrm{cover} \mathrm{that} \mathrm{in} \mathrm{next} \mathrm{half} \mathrm{cycle}. \\ \mathrm{So}, \mathrm{the} \mathrm{number} \mathrm{of} \mathrm{half} \mathrm{cycle} \mathrm{required} \mathrm{are} 3\times 2= 6. \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{If} x_1 \mathrm{and} x_2 \mathrm{be} \mathrm{the} \mathrm{displacement} \mathrm{of} \mathrm{the} \mathrm{first} \mathrm{and} \mathrm{second} \mathrm{ball} \mathrm{respectively} \mathrm{then}, \\ {x}_{\mathit{1}}\mathit{=}\frac{{v_{\mathit{1}}}^{\mathit{2}}}{2g} \mathrm{and}\mathit{ }{x}_{\mathit{2}}\mathit{=}\frac{{v_{\mathit{2}}}^{\mathit{2}}}{2g} \\ \mathrm{Now},\mathit{ }{x}_{\mathit{1}}\mathit{-}{x}_{\mathit{2}}= 15 \left(\mathrm{given}\right) \\ \mathrm{or} \frac{{v_{\mathit{1}}}^{\mathit{2}}}{2g}\mathit{-}\frac{{v_{\mathit{2}}}^{\mathit{2}}}{2g}=15 \\ \mathrm{or} \frac{4v^{\mathit{2}}}{2g}\mathit{-}\frac{v^{\mathit{2}}}{2g}\mathit{ }=15 ( \mathrm{since} 2v_{\mathit{1}}\mathit{=}{v}_{\mathit{2}} \mathrm{and} \mathrm{let}\text{'}\mathrm{s} \mathrm{say} v_{\mathit{2}}\mathit{=}v) \\ \mathrm{or} v^2=\sqrt{\frac{15\times 2\times \mathrm{g}}{3}} \mathrm{or} v=10 \mathrm{m}/\mathrm{s} \\ \mathrm{So}, v_{\mathit{1}}= 20 \mathrm{m}/\mathrm{s} \mathrm{and} v_{\mathit{2}}= 10 \mathrm{m}/\mathrm{s} . \\ \mathrm{Also}, x_{\mathit{1}}\mathit{=}\frac{{v_{\mathit{1}}}^{\mathit{2}}}{2g}=\frac{20\times 20}{20\times 10}=20 \mathrm{m} \mathrm{and} x_{\mathit{2}}=20-15=5 \mathrm{m}. \\ \mathrm{If} t_{\mathit{2}} \mathrm{is} \mathrm{the} \mathrm{time} \mathrm{taken} \mathrm{by} \mathrm{ball} 2 \mathrm{to} \mathrm{cover} \mathrm{displacement} \mathrm{of} 5 \mathrm{m} \mathrm{then}, \\ {x}_{\mathit{2}}\mathit{=}{v}_{\mathit{2}}t\mathit{-}\frac{1}{2}{g{t}_{\mathit{2}}}^{\mathit{2}} \\ \mathrm{or} 5\mathit{=}10t_{\mathit{2}}\mathit{-}5{t_{\mathit{2}}}^{\mathit{2}}\mathit{ } \\ \mathrm{or} {t_{\mathit{2}}}^{\mathit{2}}\mathit{-}2t_{\mathit{2}}\mathit{+}1\mathit{=}\mathit{0}\mathit{ } \\ \Rightarrow t_{\mathit{2}}\mathit{=}1 \mathrm{s}. \\ \mathrm{Since}\mathit{ }{t}_1= 2 \mathrm{s} \mathrm{then} \mathrm{time} \mathrm{interval} \mathrm{between} \mathrm{the} \mathrm{throws} = 2-1= 1 \mathrm{s}. \end{gathered}$$