Kinematics (motion in one dimension) Quiz-L2

Hint

(a)
\(
\begin{array}{c}
a=b t \\
\therefore \quad \int_{v_0}^v d v=\int a d t=\int_0^t b t \cdot d t \Rightarrow v=v_0+\frac{b t^2}{2}
\end{array}
\)

Further integrating the above equation w.r.t. time, we get
\(
s=v_0 t+\frac{b t^3}{6}
\)

Hint

(d)
\(
d v=a d t
\)
\(
\therefore \quad \int_2^v d v=\int_0^2\left(3 t^2+2 t+2\right) d t=\left[t^3+t^2+2 t\right]_0^2
\)
or \(\quad v=18 \mathrm{~ms}^{-1}\)

Hint

(c) \(s=t^3-6 t^2+18 t+9 \Rightarrow v=\frac{d s}{d t}=3 t^2-12 t+18\)
\(v\) is minimum or maximum at time \(t\), which can be calculated as
\(
\begin{aligned}
a & =\frac{d v}{d t}=6 t-12=0 \\
\Rightarrow \quad t & =2 \mathrm{~s}
\end{aligned}
\)
At \(t=2 \mathrm{~s}\),
\(\frac{d^2 v}{d t^2}=6>0\), i.e. \(v\) is minimum at \(t=2 \mathrm{~s}\)
\(
v_{\min }=3(2)^2-12(2)+18=6 \mathrm{~m} / \mathrm{s}
\)

Hint

(d) Uniform motion means uniform velocity or constant slope of \(s-t\) graph.

Hint

Slope of s-t graph gives instantaneous velocity
Slope \(=\tan \theta\)
Slope at \(A\) and \(D\) is very less
For \(B\) and \(C, \theta_2>\theta_1\)
\(\operatorname{tan} \theta_2>\tan \theta_1\)
Hence instantaneous velocity is maximum at \(C\).

Hint

(c) Slope of the s-t graph is velocity, \(v=\tan \theta\)
But it is valid only when angle is measured with time axis.
So, for the given graph, angle from time axis
\(
=90^{\circ}-30^{\circ}=60^{\circ}
\)
Now,
\(
v=\tan 60^{\circ}=\sqrt{3} \mathrm{~ms}^{-1}
\)

Hint

(a) \(v_i=\) slope of \(s-t\) graph \(=\tan 45^{\circ}=1 \mathrm{~ms}^{-1}\)
\(v_f=\) slope of \(s-t\) graph \(=\tan 60^{\circ}=\sqrt{3} \mathrm{~ms}^{-1}\)

Now, \(a_{\mathrm{av}}=\frac{v_f-v_i}{\Delta t}=\frac{\sqrt{3}-1}{1}=(\sqrt{3}-1)\) units

Hint

(d) Maximum acceleration means maximum change in velocity in minimum time interval.
In time interval, \(t=30 \mathrm{~s}\) to \(t=40 \mathrm{~s}\)
\(
\therefore \quad a=\frac{\Delta v}{\Delta t}=\frac{80-20}{40-30}=\frac{60}{10}=6 \mathrm{cms}^{-2}
\)

Hint

\(
\text { (b) Distance }=\text { Area under } v \text { – } t \text { graph }=A_1+A_2+A_3+A_4
\)

\(
\begin{array}{l}
=\frac{1}{2} \times 1 \times 20+(20 \times 1)+\frac{1}{2}(20+10) \times 1+(10 \times 1) \\
=10+20+15+10=55 \mathrm{~m}
\end{array}
\)

Hint

\(
\text { (c) } \text { Distance }=\text { Area of trapezium }=\frac{1}{2} \times 3.6 \times(12+8)=36 \mathrm{~m}
\)

Hint

Displacement is equal to area under the velocity time graph with proper sign.
\(\therefore\) Displacement \(=4 \times 2-2 \times 2+2 \times 2=8 \mathrm{~m}\)
Distance is equal to total area under the speed time graph.
\(\therefore\) Distance \(=4 \times 2+2 \times 2+2 \times 2=16 \mathrm{~m}\).

Hint

\(
\text { (b) } v=\frac{d x}{d t}=2 \mathrm{~ms}^{-1}=\text { constant }
\)

Hint

(a) Velocity will continuously increase (starting from rest).

Hint

(a) Velocity first decreases in upwards direction, then increases in downward direction.

Hint

\(
\text { (c) } a=\text { Slope of } v t \text { graph }=-\frac{15}{3}=-5 \mathrm{~ms}^{-2}
\)

Hint

(a) Since, total displacement is zero, hence average velocity is also zero.

Hint

(d) I is not possible because total distance covered by a particle cannot decrease with time. II is not possible because at a particular time, position cannot have two values. III is not possible because at a particular time, velocity cannot have two values. IV is not possible because speed can never be negative.

Hint

\(
\text { (d) } a=v\left(\frac{d v}{d s}\right)=(10)\left(-\frac{20}{30}\right)=-\frac{20}{3} \mathrm{~ms}^{-2}
\)

Hint

(b) For uniformly accelerated motion, \(v^2=u^2+2 a\) s, i.e.
\(v^2\) versus \(s\) graph is a straight line with intercept \(u^2\) and slope
\(2 a\). Since, intercept is non-zero, initial velocity is non-zero.

Hint

\(
\begin{array}{l}
\text { (b) } v^2=u^2+2 a s, \text { slope }=2 a=-\frac{16}{2}=-8 \mathrm{~ms}^{-2} \\
\text { or } \quad a=-4 \mathrm{~ms}^{-2}
\end{array}
\)

Hint

(a) Taking the motion from 0 to \(2 \mathrm{~s}\),
\(
\begin{array}{l}
u=0, a=3 \mathrm{~ms}^{-2}, t=2 \mathrm{~s}, v=? \\
v=u+a t=0+3 \times 2=6 \mathrm{~ms}^{-1}
\end{array}
\)

Taking the motion from \(2 \mathrm{~s}\) to \(4 \mathrm{~s}\),
\(
v=6+(-3)(2)=0 \mathrm{~ms}^{-1}
\)

Hint

\(
\text { (d) } 7.2=\frac{100}{(v+5)(5 / 18)} \text { or } v=45 \mathrm{~ms}^{-1}
\)

Hint

(a) Let positive direction of motion be from south to north.
Given,
\(
\begin{array}{l}
\mathbf{v}_A=+54 \mathrm{kmh}^{-1}=15 \mathrm{~ms}^{-1}, \\
\mathbf{v}_B=-90 \mathrm{kmh}^{-1}=-25 \mathrm{~ms}^{-1}
\end{array}
\)
\(\therefore\) The relative velocity of \(B\) w.r.t. \(A\),
\(
\mathbf{v}_{B A}=\mathbf{v}_B-\mathbf{v}_A=-25-15=-40 \mathrm{~ms}^{-1}
\)
i.e. The train \(B\) appears to \(A\) to move with a speed of \(40 \mathrm{~ms}^{-1}\) from north to south.

Hint

(c) Both the particles will fall same distance in same time interval.
So, the relative separation will remain unchanged.

Hint

\(
\text { (b) } \mathbf{v}_{A B}=\mathbf{v}_A-\mathbf{v}_B
\)

\(
\begin{aligned}
v_{A B} & =\sqrt{v_A^2+v_B^2}=5 \mathrm{kmh}^{-1} \\
\alpha & =\tan ^{-1}\left(\frac{v_A}{v_B}\right)=\tan ^{-1}\left(\frac{3}{4}\right) \\
& =37^{\circ} \text { east of north }
\end{aligned}
\)

Hint

(b) Velocity of rain is at \(30^{\circ}\) in vertical direction. So, its horizontal component is \(v_R \sin 30^{\circ}=\frac{v_R}{2}\). When man starts walking with \(10 \mathrm{kmh}^{-1}\) rain appears vertical. So, horizontal component \(\frac{v_R}{2}\) is balanced by his speed of \(10 \mathrm{kmh}^{-1}\). Thus, \(\frac{v_R}{2}=10\) or \(v_R=20 \mathrm{kmh}^{-1}\)

Alternate:

When the man is at rest with respect to the ground, the rain comes to him at an angle \(30^{\circ}\) with the vertical. This is the direction of the velocity of raindrops with respect to the ground. The situation when the man runs is shown in the figure

\(
\begin{array}{l}
\mathrm{V}_{\mathrm{rg}}=\text { velocity of rain drops with respect to ground. } \\
\mathrm{V}_{\text {man }}=\text { velocity of man }=10 \mathrm{~km} / \mathrm{h} \\
\text { Hence, } \cos \theta=\frac{10}{\mathrm{~V}_{\mathrm{rg}}} \\
\Rightarrow \cos 60^{\circ}=\frac{10}{\mathrm{~V}_{\mathrm{rg}}} \\
\Rightarrow \frac{1}{2}=\frac{10}{\mathrm{~V}_{\mathrm{rg}}} \\
\Rightarrow \mathrm{V}_{\mathrm{rg}}=20 \mathrm{~km} / \mathrm{h}
\end{array}
\)
Hence, velocity of rain drops with respect to the ground is \(20 \mathrm{~km} / \mathrm{h}\).

Hint

\(
\begin{array}{ll}
\text { (c) } \tan 60^{\circ}=\frac{v_H}{v_V} \text { or } \sqrt{3}=\frac{12}{v_V} \\
\therefore & v_V=4 \sqrt{3} \mathrm{kmh}^{-1}
\end{array}
\)

Alternate:

Let \(\vec{V}_m\) be the velocity of the running man.
Let the velocity of the rain be \(\vec{V}_r\)

From the figure, we can say \(\tan 60^{\circ}=\frac{\left|\overrightarrow{\mathrm{V}_{\mathrm{m}}}\right|}{\left|\overrightarrow{\mathrm{V}_{\mathrm{r}}}\right|}\)
\(
\begin{array}{l}
\sqrt{3}=\frac{12}{\left|\vec{V}_r\right|} \\
\therefore\left|\vec{V}_r\right|=4 \sqrt{3} \mathrm{~km} / \mathrm{h}
\end{array}
\)

Hint

(a) \(\tan \theta=\frac{v_H}{v_V}=\frac{v}{u}\)
or \(\quad \theta=\tan ^{-1}\left(\frac{v}{u}\right)\)

Hint

Shortest path that is, Width of river \(=1 \mathrm{Km}\)
Speed of boat in still water \(v_{br}=5 \mathrm{Kmhr}^{-1}\)
Time taken by boat to cross the river \(\mathrm{t}=15 \mathrm{~min}=\frac{1}{4} \mathrm{hr}\)
Let the speed of river \(=v_r\)
The vector diagram is shown below

Calculating the velocity of river:
By observing the vector diagram we can calculate the velocity along the shortest path as below:
Speed in the direction of shortest path \(=\frac{\text { shortest path }}{\text { time }}\)
And also we know that
\(
\text { time }=\frac{\text { shortest path }}{\sqrt{{v_{br}}^2-\mathrm{v_r}^2}}
\)
\(
\Rightarrow \frac{1}{4}=\frac{1}{\sqrt{25-{v_r}^2}}
\)
\(
{v_r}=3 \mathrm{Kmh}^{-1}
\)

Therefore the speed of river is equal to \(3 \mathrm{Kmh}^{-1}\).

Hint

\(
\text { (b) } \mathbf{v}_x=-v \hat{\mathbf{j}}
\)

\(
\begin{array}{ll}
\therefore \quad \mathbf{v}_Y=\mathbf{v}_{Y X}+\mathbf{v}_X=-\hat{\mathbf{i}}+v \hat{\mathbf{j}} \\
\left|\mathbf{v}_Y\right|=\sqrt{2} v
\end{array}
\)
Direction of \(\mathbf{v}_Y\) is north-west.

Hint

(a) To reach the opposite bank of the river in minimum time, the swimmer should swim at right angles to the direction of flow of river. He should swim in a direction due north

Hint

(a)

\(
\begin{array}{l}
\mathrm{v}_{\mathrm{{br}}}=5 \mathrm{~km} / \mathrm{h} \\
\mathrm{v}_{\mathrm{r}}=3 \mathrm{~km} / \mathrm{h}
\end{array}
\)
To reach the directly opposite point the man should move along the line joining the two points in other words due to the resultant velocity his body should flow perpendicular to river flow.
To do this we need to cancel out the speed of water to avoid the drifting. Suppose he moves at angle \(90+\theta\) from the direction of flow. Thus his speed component opposite to that of river is \(\mathrm{v}_{\mathrm{br}} \sin \theta\) that should cancel \(\mathrm{v}_{\mathrm{r}}\).
\(
\sin \theta=\frac{v_r}{v_{k \pi}}=\frac{3}{5}, \theta=37^{\circ}
\)
The required angle is therefore
\(
90^{\circ}+\theta=90^{\circ}+37^{\circ}=127^{\circ}
\)

Hint

(c) Let \(v\) be the speed of boatman in still water. Resultant of \(v\) and \(u\) should be along \(A B\). Components of \(\mathbf{v}_b\) (absolute velocity of boatman) along \(x\) and \(y\)-directions are,

\(
v_x=u-v \sin \theta \text { and } v_y=v \cos \theta
\)
Further, \(\tan 45^{\circ}=\frac{v_y}{v_x}\) or \(1=\frac{v \cos \theta}{u-v \sin \theta}\)
\(
\therefore \quad v=\frac{u}{\sin \theta+\cos \theta}=\frac{u}{\sqrt{2} \sin \left(\theta+45^{\circ}\right)}
\)
\(v\) is minimum at,
\(
\theta+45^{\circ}=90^{\circ} \text { or } \theta=45^{\circ} \text { and } v_{\min }=\frac{u}{\sqrt{2}}
\)

Hint

(b) \(v_r=10+15=25 \mathrm{~ms}^{-1}\)
where, \(v_r\) is relative velocity.
\(
\therefore \quad t=\frac{50+50}{25}=4 \mathrm{~s}
\)

Hint

(b) Relative acceleration \(=0\) relative velocity is \(40 \mathrm{~ms}^{-1}\) and relative separation is \(100 \mathrm{~m}\).
\(
\therefore \quad t=\frac{100}{40}=25 \mathrm{~s}
\)

Hint

(b)

\(
\begin{aligned}
\text { Average speed of a boy } & =\frac{2 v_1 v_2}{v_1+v_2}=\frac{2 \times 2.5 \times 4}{2.5+4} \\
& =\frac{20}{6.5}=\frac{40}{13} \mathrm{~km} / \mathrm{h}
\end{aligned}
\)

Hint

\(
\begin{array}{l}
\text { (a) } s_n=u+\frac{a}{2}(2 n-1) \\
\Rightarrow \quad s_3=0+\frac{4 / 3}{2}(2 \times 3-1) \Rightarrow s_3=\frac{10}{3} \mathrm{~m}
\end{array}
\)

Hint

(c) Time taken to travel first half distance, \(t_1=\frac{l / 2}{v_1}=\frac{l}{2 v_1}\) Time taken to travel second half distance, \(t_2=\frac{l}{2 v_2}\)
\(
\text { Total time }=t_1+t_2=\frac{l}{2 v_1}+\frac{l}{2 v_2}=\frac{l}{2}\left(\frac{1}{v_1}+\frac{1}{v_2}\right)
\)
We know that, \(v_{\text {av }}=\) Average speed
\(
=\frac{\text { total distance }}{\text { total time }}=\frac{l}{\frac{l}{2}\left(\frac{1}{v_1}+\frac{1}{v_2}\right)}=\frac{2 v_1 v_2}{v_1+v_2}
\)

Hint

(d) In all the given graphs, we have more than one value of speed corresponding to a single time. However, the graph must have a unique value of speed corresponding to a single time. Thus, all the three graphs are not possible.

Hint

(b) From third equation of motion, \(v^2=u^2+2 a s\)
\(
\Rightarrow \quad a=\frac{v^2-u^2}{2 \mathrm{~s}}=\frac{(20)^2-(10)^2}{2 \times 135}=\frac{300}{270}=\frac{10}{9} \mathrm{~ms}^{-2}
\)
From first equation of motion,
\(
v=u+a t \Rightarrow t=\frac{v-u}{a}=\frac{20-10}{10 / 9}=\frac{10}{10 / 9}=9 \mathrm{~s}
\)

Hint

(c) Man walks from his home to market with a speed of \(5 \mathrm{~kmh}^{-1}\). Distance \(=2.5 \mathrm{~km}\) and time
\(
=\frac{d}{v}=\frac{2.5}{5}=\frac{1}{2}=30 \mathrm{~min}
\)
and he returns back with speed of \(7.5 \mathrm{~kmh}^{-1}\) in rest a time of \(20 \mathrm{~min}\).
Distance \(=7.5 \times \frac{20}{60}=2.5 \mathrm{~km}\)
So, average speed \(=\frac{\text { total distance }}{\text { total time }}=\frac{(2.5+2.5)}{(50 / 60) \mathrm{h}}=\frac{30}{5} \mathrm{~km} / \mathrm{h}\)

Hint

(d) \(s \propto u^2\). If \(u\) becomes 3 times, then \(s\) will become 9 times, i.e. \(9 \times 20=180 \mathrm{~m}\)

Hint

(a) Distance travelled by body \(A\) in 5 th second and distance travelled by body \(B\) in 3rd second of its motion are equal.
\(
\begin{aligned}
0+\frac{a_1}{2}(2 \times 5-1) & =0+\frac{a_2}{2}(2 \times 3-1) \\
9 a_1 & =5 a_2 \Rightarrow \frac{a_1}{a_2}=\frac{5}{9}
\end{aligned}
\)

Hint

(a) \(\because\) Slope of displacement-time graph at the point \(E\) is negative. Thus, at the point \(E\), the instantaneous velocity of the particle is negative.

Hint

(d) Since \(x=1.2 t^2\), comparing it with equation of motion
\(
x=\frac{1}{2} a t^2 \text {. }
\)
Thus, the motion is uniformly accelerated.

Note: Since, slope is increasing parabolically at a constant rate hence, motion of car is uniformly accelerated.

Hint

(c) It is given that, \(v=\frac{t^2}{10}+20\)
Differentiating both sides w.r.t. time, we get
\(
\frac{d v}{d t}=\frac{2 t}{10}+0=\frac{t}{5}
\)
\(\therefore\) Acceleration, \(a=\frac{d v}{d t}=\frac{t}{5} \neq\) constant
Hence, it is a case of non-uniform acceleration.

Hint

(b) Given, \(v=\sqrt{c_1-c_2 x}\)
We know that, \(a=v \frac{d v}{d x}\)
\(
\begin{aligned}
\Rightarrow \quad a & =\sqrt{c_1-c_2 x} \cdot \frac{d}{d x} 9\left(\sqrt{c_1-c_2 x}\right) \\
& =\frac{1}{2}\left(-c_2\right)=-\frac{c_2}{2}
\end{aligned}
\)

Hint

(d) \(v_1=\frac{s}{90}, v_2=\frac{s}{60}\)

Now, \(t=\frac{s}{v_1+v_2}=\frac{s}{\frac{s}{90}+\frac{s}{60}}\)
\(
=\frac{90 \times 60}{90+60}=36 \mathrm{~s}
\)

Hint

(a) At \(t=2 \mathrm{~s} ; x_A=10 \times 2-\frac{1}{2} \times 4 \times(2)^2=12 \mathrm{~m}\) and \(x_B=20 \times 2-\frac{1}{2} \times 2 \times(2)^2=36 \mathrm{~m}\)
\(\therefore\) Distance between \(A\) and \(B\) at that instant is \(36-12=24 \mathrm{~m}\).

Hint

(a) Given, \(\sqrt{x}=t+1\)
Squaring both sides, we get
\(
x=(t+1)^2=t^2+2 t+1
\)
Differentiating it w.r.t. time \(t\), we get
\(
\frac{d x}{d t}=2 t+2
\)
Velocity, \(v=\frac{d x}{d t}=2 t+2\)

This shows velocity increases with respect to time linearly.

Hint

\(
\text { (d) Average velocity }=\frac{\text { Total displacement }}{\text { Total time }}
\)

or \(\quad 20=\frac{\frac{1}{2} \times a \times t^2+(a t)(25-2 t)+\frac{1}{2} \times a \times t^2}{25}\)
Solving this equation with \(a=5 \mathrm{~ms}^{-2}\), we get
\(
t=5 \mathrm{~s}
\)

Thus, the particle moved uniformly for \((25-2 t)\) or \(15 \mathrm{~s}\).

Hint

\(
\begin{array}{l}
\text { (d) Net displacement }=0 \text { and total distance }=O P+P Q+Q O \\
\qquad=1+\frac{2 \pi \times 1}{4}+1=\frac{14.28}{4} \mathrm{~km} \\
\text { Average speed }=\frac{14.28}{4 \times 10 / 60}=\frac{6 \times 14.28}{4}=21.42 \mathrm{~km} / \mathrm{h}
\end{array}
\)

Hint

(d) Integrate twice to convert \(a\) – \(t\) equation into \(s\) – \(t\) equation.
\(
\begin{array}{lc}
\because & a=\frac{v}{t} \Rightarrow v=a t \\
\Rightarrow & \int d v=\int a \cdot d t \Rightarrow v=\int p t \cdot d t=\frac{p t^2}{2} \\
\because & s=v \times t \Rightarrow d s=v d t \\
\Rightarrow & \int d s=\int_0^{t_1} v d t \Rightarrow s=\frac{1}{6} p t_1^3
\end{array}
\)

Hint

(c) Distance travelled by the particle is \(x=40+12 t-t^3\)

We know that, speed is defined as the rate of change of distance, i.e. \(\quad v=\frac{d x}{d t}\)
\(
v=\frac{d}{d t}\left(40+12 t-t^3\right)=0+12-3 t^2
\)
But final velocity, \(v=0\)
\(
\begin{array}{l}
\therefore \quad 12-3 t^2=0 \Rightarrow t^2=\frac{12}{3}=4 \\
\Rightarrow \quad t=2 \mathrm{~s} \\
\end{array}
\)
Hence, distance travelled by the particle before coming to rest is given by
\(
x=40+12(2)-(2)^3=40+24-8=64-8=56 \mathrm{~m}
\)

Hint

\(
\begin{array}{l}
\text { (b) Given, } s=1.2 \mathrm{~m}, v=640 \mathrm{~ms}^{-1} \text {, } \\
a=\text { ?; } u=0 ; t=\text { ? } \\
2 a s=v^2-u^2 \\
\Rightarrow \quad 2 a \times 1.2=640 \times 640 \\
\Rightarrow \quad a=\frac{8 \times 64 \times 100}{3} \\
v=u+a t \\
\Rightarrow \quad t=\frac{v}{a}=\frac{640 \times 3}{8 \times 64 \times 100}=37.5 \times 10^{-3} \mathrm{~s} \approx 40 \mathrm{~ms} \\
\end{array}
\)

Hint

(a) Here, \(u=0, a=g\)
Distance travelled in \(n\)th second is given by
\(
\begin{array}{ll}
& D_n=u+\frac{a}{2}(2 n-1) \\
\therefore \quad & D_n \propto(2 n-1) \\
\therefore \quad & D_1: D_2: D_3: D_4: D_5 \ldots=1: 3: 5: 7: 9: \ldots
\end{array}
\)

Hint

(b) The area under acceleration-time graph gives change in velocity.
As acceleration is zero at the end of \(11 \mathrm{~s}\), i.e. \(\quad v_{\max }=\) Area of \(\triangle O A B=\frac{1}{2} \times 11 \times 10=55 \mathrm{~ms}^{-1}\)

Hint

(a)

\(
\begin{aligned}
a_{\mathrm{av}} & =\frac{v_f-v_i}{\Delta t} \quad \text { (As they are in opposite direction) } \\
& =\frac{\sqrt{2 g h_f}+\sqrt{2 g h_i}}{\Delta t}=\frac{\sqrt{2 \times 10 \times 5}+\sqrt{2 \times 10 \times 10}}{0.01} \\
& =2414 \mathrm{~ms}^{-2}
\end{aligned}
\)

Alternate:

When dropped from the height \(10 \mathrm{~m}\) from rest, the velocity with which the ball hits the ground is
\(\mathrm{v_i}=\sqrt{2 \mathrm{gh}}=\sqrt{2 \times 10 \times 10}=\sqrt{196}=14.14 \mathrm{~m} / \mathrm{s}\) in vertically downward direction.
After the ball rebounds, it reaches maximum height of \(2.5 \mathrm{~m}\)
so velocity just after the rebound is
\(\mathrm{v_f}=\sqrt{2 \times 10 \times 5}=\sqrt{49}=10 \mathrm{~m} / \mathrm{s}\) in vertically upward direction.
So average acceleration is \((10-(-14.14)) / 0.01=2414 \mathrm{~m} / \mathrm{s}^2\) in vertically upward direction.

Hint

(b) Let two boys meet at point \(C\) after time ‘ \(t\) ‘ from the starting. Then, \(A C=v t, B C=v_1 t\) (see figure)

\(
(A C)^2=(A B)^2+(B C)^2 \Rightarrow v^2 t^2=a^2+v_1^2 t^2
\)
By solving, we get
\(
t=\sqrt{\frac{a^2}{v^2-v_1^2}}
\)

Hint

(b) Let \(a\) be the retardation of boggy, then distance covered by it be \(s_b\). If \(u\) is the initial velocity of boggy after detaching from train (i.e. uniform speed of train)
\(
v^2=u^2+2 a s \Rightarrow 0=u^2-2 a s
\)
\(
\Rightarrow \quad s_b=\frac{u^2}{2 a}
\)
Time taken by boggy to stop
\(
v=u+a t \Rightarrow 0=u-a t \Rightarrow t=\frac{u}{a}
\)
In this time \(t\), distance travelled by train, \(s_t=u t=\frac{u^2}{a}\) Hence, ratio \(\frac{s_b}{s_t}=\frac{1}{2}\).

Hint

(d)

Let acceleration is \(a\) and retardation is \(-2 a\).
Then, for accelerated motion,
\(
t_1=\frac{v}{a} \dots(i)
\)
For retarding motion,
\(
t_2=\frac{v}{2 a} \dots(ii)
\)
Given, \(t_1+t_2=9 \Rightarrow \frac{v}{a}+\frac{v}{2 a}=9 \Rightarrow \frac{3 v}{2 a}=9 \Rightarrow \frac{v}{a}=6\)
Hence, duration of acceleration, \(t_1=\frac{v}{a}=6 \mathrm{~s}\).

Hint

(d) Given, \(x=a e^{-\alpha t}+b e^{\beta t}\)
So, velocity, \(v=\frac{d x}{d t}=-a \alpha e^{-\alpha t}+b \beta e^{\beta t}\) \(a=\frac{d v}{d t}=a \alpha^2 e^{-\alpha t}+b \beta^2 e^{\beta t}=+\) ve all time.
\(\therefore v\) will go on increasing.

Hint

\(
\text { (b) } 1=\frac{1}{2} g t_1^2 \text { or } t_1=\sqrt{\frac{2}{g}}
\)
\(
2=\frac{1}{2} g t^2 \text { or } t=\sqrt{\frac{4}{g}}
\)
But
\(
t_2=t-t_1=\sqrt{\frac{4}{g}}-\sqrt{\frac{2}{g}}
\)
\(
\begin{aligned}
\therefore \quad \frac{t_1}{t_2} & =\frac{\sqrt{2 / g}}{\sqrt{4 / g}-\sqrt{2 / g}}=\frac{\sqrt{2}}{2-\sqrt{2}} \\
& =\frac{\sqrt{2}(2+\sqrt{2})}{2}=(\sqrt{2}+1)
\end{aligned}
\)

Hint

(c) \(t=\sqrt{\frac{x+a}{b}}\) or \((x+a)=b t^2\) or \(x=-a+b t^2\) Comparing this equation with general equation of uniformly accelerated motion, \(s=s_i+u t+\frac{1}{2} a t^2\) we see that \(s_i=-a, u=0\) and acceleration \(=2 b\).

Hint

(b) Given, acceleration, \(a=6 t+5\)
\(
\therefore \quad a=\frac{d v}{d t}=6 t+5, d v=(6 t+5) d t
\)
Integrating it, we have
\(
\int_0^v d v=\int_0^t(6 t+5) d t
\)
\(v=3 t^2+5 t+C\), where \(C\) is a constant of integration.
When \(t=0, v=0\), so \(C=0\)
\(
\therefore \quad v=\frac{d s}{d t}=3 t^2+5 t \text { or } d s=\left(3 t^2+5\right) d t
\)
Integrating it within the conditions of motion, i.e. as \(t\) changes from 0 to \(2 \mathrm{~s}, s\) changes from 0 to \(s\), we have
\(
\begin{aligned}
& \int_0^s d s =\int_0^2\left(3 t^2+5 t\right) d t \\
\therefore & s =\left[t^3+\frac{5}{2} t^2\right]_0^2=8+10=18 \mathrm{~m}
\end{aligned}
\)

Hint

(a) Given, \(x=\frac{1}{t+5}\)
Differentiating both sides w.r.t. \(t\), we get
\(
\frac{d x}{d t}=-\frac{1}{(t+5)^2} \times 1
\)
\(\therefore\) Velocity, \(v=\frac{d x}{d t}=-\frac{1}{(t+5)^2} \dots(i)\)
Again, differentiating both sides w.r.t. \(t\), we get
\(
\frac{d v}{d t}=\frac{1 \times 2}{(t+5)^3}(1)=\frac{2}{(t+5)^3} \dots(ii)
\)
Now, from Eqs. (i) and (ii), we get
\(
\frac{d v}{d t}=-2 v \times \frac{1}{(t+5)}=(-2 v)(\sqrt{v})=-2(v)^{3 / 2}
\)
\(\therefore\) Acceleration, \(a=\frac{d v}{d t}=-2 v^{3 / 2} \Rightarrow a \propto(\text { velocity })^{3 / 2}\)

Hint

(b) \(v=\frac{d x}{d t}=32-6 t^2, v=0\) at \(t=2.30 \mathrm{~s}\)
\(
a=\frac{d v}{d t}=-12 t
\)

At \(2.30 \mathrm{~s}\),
\(
a=-27.6 \mathrm{~ms}^{-2}
\)

Hint

(b) For \(P, 30=15+a t\) or at \(=15 \mathrm{~ms}^{-1}\)

For \(Q, v=20-a t\) or \(v=20-15=5 \mathrm{~ms}^{-1}\)

Hint

(c) Let the man will catch the bus after \(t\) second. So, he will cover distance \(u t\).
Similarly, distance travelled by the bus will be \(\frac{1}{2} a t^2\).
For the given condition,
\(
\begin{aligned}
& u t=45+\frac{1}{2} a t^2=45+1.25 t^2 \quad\left(\text { As, } a=2.5 \mathrm{~ms}^{-2}\right) \\
\Rightarrow \quad & u=\frac{45}{t}+1.25 t
\end{aligned}
\)
To find the minimum value of \(u\)
\(
\frac{d u}{d t}=0 \Rightarrow \frac{d u}{d t}=\frac{-45}{t^2}+1.25=0
\)

So, we get \(t=6 \mathrm{~s}\), then \(u=\frac{45}{6}+(1.25 \times 6)=15 \mathrm{~ms}^{-1}\)

Hint

\(
\begin{array}{l}
\text { (b) } x^2=t^2+1 \text { or } 2 x \cdot \frac{d x}{d t}=2 t \text { or } x \cdot \frac{d x}{d t}=t \\
\therefore \quad v=\frac{d x}{d t}=\frac{t}{x}=\frac{t}{\sqrt{t^2+1}} \\
\quad a=\frac{d v}{d t}=\frac{d^2 x}{d t^2}=\frac{\sqrt{t^2+1}-\frac{t^2}{\sqrt{t^2+1}}}{\left(t^2+1\right)}=\frac{1}{\left(t^2+1\right)^{3 / 2}}=\frac{1}{x^3}
\end{array}
\)

Hint

(b) Average velocity in uniformly accelerated motion is given by
\(
v_{\mathrm{av}}=\frac{s}{t}=\frac{u t+\frac{1}{2} a t^2}{t}=u+\frac{1}{2} a t
\)
Now, \(v_1=u+\frac{1}{2} a t_1, v_2=\left(u+a t_1\right)+\frac{1}{2} a t_2\)
\(
\begin{array}{l}
\text { and } v_3=u+a\left(t_1+t_2\right)+\frac{1}{2} a t_3 \\
\therefore\left(v_1-v_2\right):\left(v_2-v_3\right)=\left(t_1+t_2\right):\left(t_2+t_3\right)
\end{array}
\)

Hint

(d) Motion is first uniformly accelerated in positive direction, then it is uniform in negative direction.

Hint

\(
\text { (c) } h=-5 t+5 t^2 \text { and } 2 h=5 t+5 t^2
\)

From these two equations, we get \(t=3 \mathrm{~s}\) and \(h=\) distance between \(P\) and \(Q=30 \mathrm{~m}\).

Hint

(b) Let height of Minaret be \(H\) and body take time \(T\) to fall from top to bottom.

\(
H=\frac{1}{2} g T^2 \dots(i)
\)
In last \(2 \mathrm{~s}\), body travels distance of \(40 \mathrm{~m}\), so in \((T-2)\) second, distance travelled \(=(H-40) \mathrm{m}\)
\(
\Rightarrow \quad(H-40)=\frac{1}{2} g(T-2)^2 \dots(ii)
\)
By solving Eqs. (i) and (ii), we get
\(
T=3 \mathrm{~s} \text { and } H=45 \mathrm{~m}
\)

Hint

(a) Given acceleration-time graph is shown in figure.

Let \(a(t=0)=a_0=\) constant
From the graph, we have
\(
\begin{aligned}
\frac{t}{T}+\frac{a}{a_0} & =1 \\
\Rightarrow \quad a_0 t+a T & =a_0 T \dots(i)
\end{aligned}
\)
Also, we know that,
\(
a=\frac{d v}{d t} \dots(ii)
\)
From Eqs. (i) and (ii), we get
\(
\frac{a_0 T-a_0 t}{T}=\frac{d v}{d t}
\)
Integrating both sides, we get
\(
\begin{aligned}
\quad \int_0^t\left(a_0 T-a_0 t\right) d t & =T \int_0^v d v \\
\Rightarrow \quad a_0 T t-\frac{a_0 t^2}{2} & =T v \Rightarrow v=a_0 t-\frac{a_0}{2 T} t^2
\end{aligned}
\)

Therefore, \(v\)-t graph must be parabolic.
Now, at \(t=0, v=0\)
\(
t=T, v=\frac{a_0 T}{2}=\text { constant }
\)
These conditions are satisfied by graph (a).

Hint

(a) Given line has positive intercept but negative slope. So, its equation can be written as
\(
v=-m x+v_0\left(\text { where, } m=\tan \theta=\frac{v_0}{x_0}\right) \dots(i)
\)
By differentiating with respect to time, we get
\(
\frac{d v}{d t}=-m \frac{d x}{d t}=-m v
\)
Now, substituting the value of \(v\) from Eq. (i), we get
\(
\begin{aligned}
\frac{d v}{d t} & =-m\left(-m x+v_0\right)=m^2 x-m v_0 \\
\therefore \quad a & =m^2 x-m v_0
\end{aligned}
\)
i.e. The graph between \(a\) and \(x\) should have positive slope but negative intercept on \(a\)-axis. So, graph (a) is correct.

Hint

(b) \(\frac{d x}{d t}=v=-1-2 t\)
Comparing with \(v=u+a t\), we get
\(
u=-1 \mathrm{~ms}^{-1} \text { and } a=-2 \mathrm{~ms}^{-2} \text {. }
\)
At \(t=0, x=1 \mathrm{~m}\) Then, \(u\) and \(a\) both are negative. Hence, \(x\)-coordinate of particle will go on decreasing.

Hint

(b) In graph (b) for one value of displacement, there are two different points of time. Hence, for one time, the average velocity is positive and for other time is equivalent negative.
As there are opposite velocities in the interval 0 to \(T\), hence average velocity can vanish in (b). This can be seen in the figure alongside. Here, \(O A=B T\) (same displacement) for two different points of time.

Hint

(a) As the lift is coming in downward directions, displacement will be negative. We have to see whether the motion is accelerating or retarding.
We know that due to downward motion, displacement will be negative. When the lift reaches 4 th floor and is about to stop, hence motion is retarding in nature, hence \(x<0 ; a>0\). As displacement is in negative direction, velocity will also be negative, i.e. \(v<0\).

Hint

(b) Given, \(\quad x=(t-2)^2\)
Velocity, \(\quad v=\frac{d x}{d t}=\frac{d}{d t}(t-2)^2=2(t-2) \mathrm{ms}^{-1}\)
Acceleration, \(a=\frac{d v}{d t}=\frac{d}{d t}[2(t-2)]=2(1-0)=2 \mathrm{~ms}^{-2}\)
When \(t=0 ; v=-4 \mathrm{~ms}^{-1}, t=2 \mathrm{~s} ; v=0 \mathrm{~ms}^{-1}\)
\(
t=4 \mathrm{~s} ; v=4 \mathrm{~ms}^{-1}
\)
\(v\)-t graph is shown in figure below.

\(
\begin{aligned}
\text { Distance travelled } & =\text { Area under the } v \text { – } t \text { graph } \\
& =\mid \text { Area } O A C \mid+ \text { Area } A B D \\
& =\left|\frac{4 \times 2}{2}\right|+\frac{1}{2} \times 2 \times 4=8 \mathrm{~m}
\end{aligned}
\)

Hint

(d) Let height of tower is \(h\) and body takes \(t\) time to reach to ground when it fall freely.
\(
\therefore \quad h=\frac{1}{2} g t^2 \dots(i)
\)
In last second, i.e. t-th second body travels \(=0.36 \mathrm{~h}\)
It means in rest of the time, \(i . e\) in \((t-1)\) second, it travels
\(
=h-0.36 h=0.64 h
\)
Now, applying equation of motion for \((t-1)\) second
\(
0.64 h=\frac{1}{2} g(t-1)^2 \dots(ii)
\)

From Eqs. (i) and (ii), we get
\(
t=5 \mathrm{~s} \text { and } h=125 \mathrm{~m}
\)

Hint

\(
\begin{array}{l}
\text { (a) Acceleration, } f=f_0\left(1-\frac{t}{T}\right) \\
\frac{d v}{d t}=f_0 \cdot\left(1-\frac{t}{T}\right) \quad\left(\because f=\frac{d v}{d t}\right) \\
\end{array}
\)
\(
\begin{aligned}
d v & =f_0 \cdot\left(1-\frac{t}{T}\right) d t \Rightarrow \int d v=\int f_0\left(1-\frac{t}{T}\right) d t \\
v & =f_0 t-f_0 \frac{t^2}{2 T}+C \dots(i)
\end{aligned}
\)
where, \(C\) is constant.
When \(t=0, v=0\) thus from Eq. (i), \(C=0\)
\(
v=f_0 t-\frac{f_0}{T} \cdot \frac{t^2}{2} \dots(ii)
\)
As, \(\quad f=f_0\left(1-\frac{t}{T}\right)\)
When \(f=0\)
\(
\therefore \quad f_0\left(1-\frac{t}{T}\right)=0 ; f_0 \neq 0 \Rightarrow t=T
\)
Putting \(t=T\) in Eq. (ii), we get
\(
v=f_0 T-\frac{f_0}{T} \cdot \frac{T^2}{2}=\frac{f_0 T}{2}
\)

Hint

(d) Relative to lift,
\(
\begin{array}{l}
u_r=0, a_r=(9.8+1.2)=11 \mathrm{~ms}^{-2} \\
s_r=\frac{1}{2} a r^2 \Rightarrow 2.7=\frac{1}{2} \times 11 \times t^2 \Rightarrow t=\sqrt{\frac{5.4}{11}} \mathrm{~s}
\end{array}
\)

Hint

(a) After bailing out from point \(A\), parachutist falls freely under gravity. The velocity acquired by it will be \(v\).
From \(v^2=u^2+2 a s=0+2 \times 9.8 \times 50=980\)
\(
\text { (As, } \left.u=0, a=9.8 \mathrm{~ms}^{-2}, s=50 \mathrm{~m}\right)
\)
At point \(B\), parachute opens and it moves with retardation of \(2 \mathrm{~ms}^{-2}\) and reach at ground (point \(C\) with velocity of \(3 \mathrm{~ms}^{-1}\) ). For the part \(B C\) by applying the equation \(v^2=u^2+2 a s\)
\(
v=3 \mathrm{~ms}^{-1}, u=\sqrt{980} \mathrm{~ms}^{-1}, a=-2 \mathrm{~ms}^{-2}, s=h
\)
\(
\begin{aligned}
(3)^2 & =(\sqrt{980})^2+2 \times(-2) \times h \\
9 & =980-4 h \\
h & =\frac{980-9}{4}=\frac{971}{4}=242.7 \cong 243 \mathrm{~m}
\end{aligned}
\)
So, the total height by which parachutist bails out
\(
=50+243=293 \mathrm{~m}
\)

Hint

(c) Initial relative velocity \(=v_1-v_2\), final relative velocity \(=0\)
From
\(
\begin{array}{ll}
\text { From } & v^2=u^2-2 a s \Rightarrow 0=\left(v_1-v_2\right)^2-2 \times a \times s \\
\Rightarrow & s=\frac{\left(v_1-v_2\right)^2}{2 a}
\end{array}
\)
If the distance between two cars is \(s\), then collision will take place. To avoid collision \(d>s\)
\(
\therefore \quad d>\frac{\left(v_1-v_2\right)^2}{2 a}
\)
where, \(d=\) actual initial distance between two cars.

Hint

(b) Time taken by first drop to reach the ground, \(t=\sqrt{\frac{2 h}{g}}\) \(\Rightarrow \quad t=\sqrt{\frac{2 \times 5}{10}}=1 \mathrm{~s}\)
As the water drops fall at regular intervals from a tap, therefore time difference between any two drops \(=\frac{1}{2} \mathrm{~s}\).
In this given time, distance of second drop from the tap
\(
=\frac{1}{2} g\left(\frac{1}{2}\right)^2=\frac{10}{8}=1.25 \mathrm{~m}
\)

Its distance from the ground \(=5-1.25=3.75 \mathrm{~m}\)

Hint

(d)

\(
\begin{aligned}
A & \Rightarrow \frac{u^2}{4}-u^2=-2 g h_1 \\
B & \Rightarrow \frac{u^2}{9}-u^2=-2 g h_2 \\
C & \Rightarrow \frac{u^2}{16}-u^2=-2 g h_3
\end{aligned}
\)
\(
\therefore \quad A B=\frac{u^2}{2 g}\left(\frac{8}{9}-\frac{3}{4}\right)=\frac{u^2}{2 g} \cdot \frac{5}{36}
\)
\(
B C=\frac{u^2}{2 g}\left(\frac{15}{16}-\frac{8}{9}\right)=\frac{u^2}{2 g} \cdot \frac{7}{144}
\)
\(
\frac{A B}{B C}=\frac{5}{36} \times \frac{144}{7}=\frac{20}{7}
\)

Hint

(a) For the given condition, initial height \(h=d\) and velocity of the ball is zero. When the ball moves downward, its velocity increases and it will be maximum, when the ball hits the ground and just after the collision, it becomes half and in opposite direction. As the ball moves upwards, its velocity again decreases and becomes zero at height \(d / 2\). This explanation match with graph (a).

Hint

(c) The position \(x\) of a particle w.r.t. time \(t\) along \(X\)-axis,
\(
x=9 t^2-t^3 \dots(i)
\)
Differentiating Eq. (i) w.r.t. time, we get speed, i.e.
\(
\begin{array}{l}
v=\frac{d x}{d t}=\frac{d}{d t}\left(9 t^2-t^3\right) \\
v=18 t-3 t^2 \dots(ii)
\end{array}
\)
Again, differentiating Eq. (ii) w.r.t. time, we get acceleration, i.e.
\(
\begin{aligned}
a & =\frac{d v}{d t}=\frac{d}{d t}\left(18 t-3 t^2\right) \\
\Rightarrow \quad a & =18-6 t \dots(iii)
\end{aligned}
\)
Now, when speed of particle is maximum, its acceleration is zero.
\(
\begin{array}{rlrl}
a =0 \\
\Rightarrow & 18-6 t & =0 \\
\Rightarrow & & t =3 \mathrm{~s}
\end{array}
\)
Putting in Eq. (i), we obtain position of particle at that time
\(
\begin{aligned}
x & =9(3)^2-(3)^3=9(9)-27 \\
& =81-27=54 \mathrm{~m}
\end{aligned}
\)

Hint

(c) Relative velocity of car \(A\) w.r.t. ground \(\mathbf{v}_{A g}=30 \mathrm{~km} / \mathrm{h}\)
Relative velocity of car \(B\) w.r.t. ground \(\mathbf{v}_{B g}=40 \mathrm{~km} / \mathrm{h}\)
\(\therefore\) Relative velocity of \(A\) w.r.t. \(B\),
\(
\begin{aligned}
& \mathbf{v}_{A B}=\mathbf{v}_{A g}-\mathbf{v}_{B g}=\mathbf{v}_{A g}+\left(-\mathbf{v}_{B g}\right) \\
& \left|\mathbf{v}_{A B}\right|=\sqrt{(30)^2+(40)^2}=50 \mathrm{~km} / \mathrm{h} \\
\therefore & \tan \theta=\frac{30}{40}=\frac{3}{4} \\
\Rightarrow & \quad \theta=\tan ^{-1}(3 / 4) \mathrm{N} \text { of west }
\end{aligned}
\)

Hint

(d)

Relative velocity of man w.r.t. ground \(\mathbf{v}_{m g}=3 \mathrm{~m} / \mathrm{s}\)
Relative velocity of rain w.r.t. ground, \(\mathbf{v}_{\mathrm{rg}}=4 \mathrm{~m} / \mathrm{s}\)
Relative velocity of rain w.r.t. \(\operatorname{man}, \mathbf{v}_{r m}=\) ?
\(
\mathbf{v}_{m m}=\mathbf{v}_{r g}-\mathbf{v}_{m g}=\mathbf{v}_{r g}+\left(-\mathbf{v}_{m g}\right)
\)
\(
\begin{aligned}
\mathbf{v}_{\mathrm{rm}} & =5 \mathrm{~m} / \mathrm{s} \\
\tan \alpha & =\frac{3}{4}, \quad \tan \beta=\frac{4}{3}
\end{aligned}
\)
\(\mathbf{v}_{m m}=5 \mathrm{~m} / \mathrm{s}\) at \(\tan ^{-1}\left(\frac{3}{4}\right)\) with vertical or at \(\tan ^{-1}\left(\frac{4}{3}\right)\) with horizontal.

Hint

where, \(\mathbf{v}_{b g}\) is the velocity of boat w.r.t. ground and \(\mathbf{v}_{s g}\) is the velocity of ship w.r.t. ground.
Relative velocity of boat w.r.t. ship \(\mathbf{v}_{b s}\) is along north. \(\mathbf{v}_{b s}\) is resultant of \(\mathbf{v}_{b g}\) and \(\mathbf{v}_{s g}\) in opposite direction.
\(
\begin{aligned}
v \cos 30^{\circ} & =v_{b s} \\
v \sin 30^{\circ} & =15 \Rightarrow v=30 \mathrm{~km} / \mathrm{h}
\end{aligned}
\)

Hint

\(u: v_{r g}\) (velocity of river w.r.t. ground)
\(v: v_{m r}\) (velocity of man w.r.t. river)
Along the flow, \(v_{m g}=v+u, 3=\frac{d}{v+u} \dots(i)\)
where, \(v_{m g}\) is velocity of man w.r.t. ground.
Opposite to the flow,
\(
\begin{aligned}
v_{m g} & =v-u, 6=\frac{d}{v-u} \dots(ii) \\
\Rightarrow \quad t_0 & =\frac{d}{v}=4 \mathrm{~h}
\end{aligned}
\)

Hint

\(
\begin{array}{rlrl}
y & \text { displacement } & =\text { velocity } \times \text { time } \\
\Rightarrow \quad 500 & =10 t \\
\Rightarrow \quad & t =50 \mathrm{~s}
\end{array}
\)

Hint

The condition of both the ships are shown in figure \((A)\) and figure \((B)\) shows the condition of ship 2 when it is always due north to the ship 1.

So, east component of both ships must be same.
\(
\begin{array}{l}
\therefore \mathrm{v}_{\mathrm{s} 2} \sin 30^{\circ}=\mathrm{v}_{\mathrm{s} 1} \\
\mathrm{v}_{\mathrm{s} 2}=\frac{10}{\frac{1}{2}}=20 \mathrm{~km} / \mathrm{hr}
\end{array}
\)

Hint

If the man starts walking on the trolley in the forward direction then whole system will move in backward direction with same momentum. Trolley will move back with a velocity \(v^{\prime}=\frac{m v}{M+m}\)
In 4 sec the net distance of the man moved \(=x=v t-v^{\prime} t=1 \times 4-\frac{320 \times 1}{320+80}=3.2 \mathrm{~m}\)

Hint

It is clear from the diagram that the shortest distance between the ship A and \(\mathrm{B}\) is \(\mathrm{PQ}\).
Here, \(\operatorname{Sin} 45^{\circ}=\frac{\mathrm{PQ}}{\mathrm{OQ}}\)
\(
\Longrightarrow \mathrm{PQ}=100 \times \frac{1}{\sqrt{2}}=50 \sqrt{2} \mathrm{~m}
\)
Also,
\(
\mathrm{V}_{\mathrm{AB}}=\sqrt{\mathrm{V}_{\mathrm{A}}^2+\mathrm{V}_{\mathrm{B}}^2}=\sqrt{10^2+10^2}=10 \sqrt{2} \mathrm{~km} / \mathrm{h}
\)
So, the time taken to reach shortest path is
\(
\mathrm{t}=\frac{\mathrm{PQ}}{\mathrm{V}_{\mathrm{AB}}}=\frac{50 \sqrt{2}}{10 \sqrt{2}}=5 \mathrm{~h}
\)

Hint

Time to cross river \((t)=\frac{A B}{V_{m r} \sin \theta}=\frac{0.4}{5 \sin \theta}\)

Drift:

\(
\begin{array}{l}
B C=\left(v_{m r} \cos \theta+v_r\right) t \\
\Rightarrow 0.4=(5 \cos \theta+1) \times \frac{0.4}{5 \sin \theta} \\
\Rightarrow 5 \sin \theta-5 \cos \theta=1 \\
\Rightarrow 25 \sin ^2 \theta+25 \cos ^2 \theta-50 \sin \theta \cos \theta=1 \\
\Rightarrow 25 \sin 2 \theta=24 \\
\Rightarrow \sin 2 \theta=\frac{24}{25} \\
\Rightarrow \theta=53^{\circ}
\end{array}
\)

Hint

(b) Relative speed of \(Q\) with respect to \(P\) is given by \((10-2) \mathrm{m} / \mathrm{s}=8 \mathrm{~m} / \mathrm{s}\)
So, if we assume \(P\) is at rest, then we can say that \(Q\) is moving towards \(P\) with speed \(8 \mathrm{~m} / \mathrm{s}\)
To reach \(P, Q\) will have to cover distance of \(B C+C D+D A=3 \times 8=24 \mathrm{~m}\)
\(
\text { time }=\frac{24}{8}=3 \text { seconds }
\)

Hint

\(
\begin{array}{l}
\text { Given } \overrightarrow{A B}=\text { velocity of boat }=8 \mathrm{~km} / \mathrm{hr} \\
\mathrm{\vec{AC}}=\text { Resultant velocity of boat }=10 \mathrm{~km} / \mathrm{hr} \\
\overrightarrow{\mathrm{BC}}=\text { Velocity of river }=\sqrt{\mathrm{AC}^2-\mathrm{AB}^2} \\
=\sqrt{(10)^2-(8)^2}=6 \mathrm{~km} / \mathrm{hr} .
\end{array}
\)