Past NEET Papers

Hint

As \(t_{1} = \frac{x}{v_{1}}\) and \(t_{2}=\frac{x}{v_{2}}\)
\(
\therefore \quad v=\frac{x+x}{t_{1}+t_{2}}=\frac{2 x}{\frac{x}{v_{1}}+\frac{x}{v_{2}}}=\frac{2 v_{1} v_{2}}{v_{1}+v_{2}}
\)
\(
\therefore \quad \frac{2}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}
\)

Hint

Speed of walking \(=\frac{h}{t_{1}}=v_{1}\)
Speed of escalator \(=\frac{h}{t_{2}}=v_{2}\)
Time taken when she walks over running escalator
\(
\begin{aligned}
&\Rightarrow \quad t=\frac{h}{v_{1}+v_{2}} \\
&\Rightarrow \quad \quad \frac{1}{t}=\frac{v_{1}}{h}+\frac{v_{2}}{h}=\frac{1}{t_{1}}+\frac{1}{t_{2}} \\
&\Rightarrow \quad t=\frac{t_{1} t_{2}}{t_{1}+t_{2}}
\end{aligned}
\)

Hint

Let the total distance covered by the particle be \(2 s\). Then
\(
v_{a v}=\frac{2 s}{\frac{s}{v_{1}}+\frac{s}{v_{2}}}=\frac{2 v_{1} v_{2}}{v_{1}+v_{2}}
\)

Hint

Average speed
\(
=\frac{\text { total distance travelled }}{\text { total time taken }}
\)
Let \(s\) be the distance from \(X\) to \(Y\).
\(\therefore\) Average speed \(=\frac{s+s}{t_{1}+t_{2}}=\frac{2 s}{\frac{s}{v_{u}}+\frac{s}{v_{d}}}\) \(=\frac{2 v_{u} v_{d}}{v_{d}+v_{u}}\)

Hint

Given,
Initial velocity \(u=0\), time \(t=20 \mathrm{~s}\)
Final velocity \(v=144 \mathrm{~km} / \mathrm{h}=40 \mathrm{~m} / \mathrm{s}\)
From 1st equation of motion,
\(
\Rightarrow \quad \begin{aligned}
v &=u+a t \\
a &=\frac{v-u}{t}=\frac{40-0}{20}=2 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)
Now, from 2nd equation of motion,
\(
\begin{aligned}
&\text { distance covered, } s=u t+\frac{1}{2} \text { at }{ }^{2} \\
&=0+\frac{1}{2} \times 2 \times(20)^{2}=400 \mathrm{~m}
\end{aligned}
\)

Hint

\(\text { Average speed }=\frac{\text { Total distance }}{\text { Total time }}\)

\(\begin{aligned}
\text { Average speed } &=\frac{\mathrm{s}}{\frac{\mathrm{s} / 3}{10}+\frac{\mathrm{s} / 3}{20}+\frac{\mathrm{s} / 3}{60}} \\
&=\frac{\mathrm{s}}{\mathrm{s} / 18}=18 \mathrm{~km} / \mathrm{h}
\end{aligned}\)

Hint

Average speed \(=\frac{\text { Total distance }}{\text { Total time }}\)
Let \(t_{1}, t_{2}\) be time taken during first-half and second-half respectively.
So, \(\quad t_{1}=\frac{100}{40} \mathrm{~s}\)
and \(\mathrm{t}_{2}=\frac{100}{v} \mathrm{~s}\)
So, according to average speed formula
\(
\begin{aligned}
48 &=\frac{200}{\left(\frac{100}{40}\right)+\left(\frac{100}{v}\right)} \\
\text { or } \quad \frac{1}{40}+\frac{1}{v} &=\frac{2}{48}=\frac{1}{24} \\
\text { or } \quad \frac{1}{v} &=\frac{2}{120}=\frac{1}{60} \\
\Rightarrow \quad \quad v &=60 \mathrm{~km} / \mathrm{h}
\end{aligned}
\)

Hint

Total distance \(=\mathrm{s}\);
Total time taken
\(
=\frac{\mathrm{s} / 2}{40}+\frac{\mathrm{s} / 2}{60}=\frac{5 \mathrm{~s}}{240}=\frac{\mathrm{s}}{48}
\)
\(\therefore\) Average speed \(=\frac{\text { total distance }}{\text { total time }}\)
\(
\begin{aligned}
&=\frac{\mathrm{s}}{\mathrm{s} / 48}=48 \mathrm{~km} / \mathrm{h} \\
\text {Alternate way}, &v_{a v}=\frac{2 v_{1} v_{2}}{v_{1}+v_{2}}=\frac{2 \times 40 \times 60}{40+60}=48 \mathrm{~km} / \mathrm{h}
\end{aligned}
\)

Hint

Given :Velocity \(v=\mathrm{At}+\mathrm{Bt}^{2}\)
\(\Rightarrow \frac{d x}{d t}=\mathrm{At}+\mathrm{Bt}^{2}\)
By integrating we get distance travelled by the particle between \(1 \mathrm{~s}\) and \(2 \mathrm{~s}\),
\(
\begin{aligned}
&\Rightarrow \int_{0}^{x} d x=\int_{1}^{2}\left(A t+B t^{2}\right) d t \\
&x=\frac{A}{2}\left(2^{2}-1^{2}\right)+\frac{B}{3}\left(2^{3}-1^{3}\right)=\frac{3 A}{2}+\frac{7 B}{3}
\end{aligned}
\)

Hint

Given, \(v(x)=b x^{-2 n}\)
\(
\begin{aligned}
a=\frac{d v}{d t} &=\frac{d v}{d x} \cdot \frac{d x}{d t} \\
&=v \cdot \frac{d v}{d x}
\end{aligned}
\)
So, \(\frac{d v}{d x}=-2 n b x^{-2 n-1}\)
Acceleration of the particle as function of \(x\),
\(
\begin{aligned}
a &=v \frac{d v}{d x}=b x^{-2 n}\left\{b(-2 n) x^{-2 n-1}\right\} \\
&=-2 n b^{2} x^{-4 n-1}
\end{aligned}
\)

Hint

\(\because t=\sqrt{x}+3\)
\(\Rightarrow \sqrt{x}=t-3 \Rightarrow x=(t-3)^{2}\)
\(v=\frac{d x}{d t}=2(t-3)=0\)
\(\Rightarrow t=3\)
\(
\therefore x=(3-3)^{2}
\)
\(
\Rightarrow x=0
\)

Hint

\(\begin{aligned}
&\vec{v}=\vec{u}+\vec{a} t \\
&v=(2 \hat{i}+3 \hat{j})+(0.3 \hat{i}+0.2 \hat{j})^{\prime} 10=5 \hat{i}+5 \hat{j} \\
&|\vec{v}|=\sqrt{5^{2}+5^{2}} \\
&|\vec{v}|=5 \sqrt{2}
\end{aligned}\)

Hint

\(x=8+12 t-t^{3}\)
The final velocity of the particle will be zero because it retarded.
Now the retardation
\(
\begin{aligned}
&a=\frac{d v}{d t}=0-6 t \\
&a[t=2]=-12 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
\)
\(
\text { retardation }=12 \mathrm{~m} / \mathrm{s}^{2}
\)

Hint

Average acceleration
\(
\begin{aligned}
n &=\frac{\text { Change in velocity }}{\text { Total time }} \\
a &=\frac{\left|\mathbf{v}_{f}-\mathbf{v}_{i}\right|}{\Delta t}=\frac{\sqrt{30^{2}+40^{2}}}{10} \\
&=\frac{\sqrt{900+1600}}{10}=5 \mathrm{~ms}^{-2}
\end{aligned}
\)

Hint

Given, initial velocity \((u)=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}\)
Final velocity \((v)=\) ?
Acceleration \((a)=(0.4 \hat{\mathbf{i}}+0.3 \hat{\mathbf{j}})\)
Time \((t)=10 \mathrm{~s}\)
From first equation of motion, \(v=u+\) at
\(
\begin{aligned}
&v=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+10(0.4 \hat{\mathbf{i}}+0.3 \hat{\mathbf{j}}) \\
&v=7 \hat{\mathbf{i}}+7 \hat{\mathbf{j}} \Rightarrow|v|=7 \sqrt{2}
\end{aligned}
\)

Hint

distance \(x=\frac{1}{t+5}\)
\(\therefore \quad\) velocity \(v=\frac{d x}{d t}=\frac{-1}{(t+5)^{2}}\)
\(\therefore \quad\) acceleration \(a=\frac{d^{2} x}{d t^{2}}=\frac{2}{(t+5)^{3}}=2 x^{3}\)
Therefore, \(v^{3 / 2}=-(t+5)^{-3}\)
\(\mathrm{So}, a \propto v^{3 / 2}\)

Hint

Since the body starts from rest \(u=0\)
\(
\therefore \quad s=\frac{1}{2} a t^{2}
\)
Now, \(\quad s_{1}=\frac{1}{2} a(10)^{2}\)
and \(\quad s_{2}=\frac{1}{2} a(20)^{2}\)
Dividing \(s_{1}  \text and  s_{2}\), we get
\(
\begin{aligned}
\frac{s_{1}}{s_{2}} &=\frac{(10)^{2}}{(20)^{2}} \\
\Rightarrow \quad s_{2} &=4 s_{1}
\end{aligned}
\)

Hint

Distance travelled in \(n^{\text {th }}\) second is given by
\(
s_{n}=u+\frac{1}{2} a(2 n-1)
\)
Here, \(u=0, a=\frac{4}{3}\)
\(
\therefore \quad s_{3}=0+\frac{1}{2} \times \frac{4}{3} \times(6-1)=\frac{10}{3} \mathrm{~m}
\)

Hint

Maximum velocity point means the point at which \(\frac{d x}{d t}\) i.e. the slope of the graph is maximum.
At point \(C\), slope is maximum.

Hint

\(\begin{aligned}
&\text { Using } v^{2}-u^{2}=2 a s \\
&\begin{array}{l}
(20)^{2}-(10)^{2}=2 \times a \times 135 \\
\Rightarrow \frac{300}{270}=a=\frac{10}{9} \\
\text { Now, using } v-u=a t \\
20-10=\frac{10}{9} \times t \\
\Rightarrow \quad t=9 \mathrm{~s}
\end{array}
\end{aligned}\)

Hint

Speed \(v=\frac{d x}{d t}=\frac{d}{d t}\left(9 t^{2}-t^{3}\right)=18 t-3 t^{2}\)
For maximum speed its acceleration should be zero,
Acceleration \(\frac{d v}{d t}=0 \Rightarrow 18-6 t=0 \Rightarrow t=3\)
At \(t=3\) its speed is max
\(
\Rightarrow x_{\max }=81-27=54 \mathrm{~m}
\)

Hint

Acceleration \(f=f_{0}\left(1-\frac{t}{T}\right)\)
or, \(\frac{d v}{d t}=f_{0}\left(1-\frac{t}{T}\right)\)
If \(f=0\), then
\(
0=f_{0}\left(1-\frac{t}{T}\right) \Rightarrow t=T
\)
Hence, particle’s velocity in the time interval \(t=0\) and \(t=T\) is given by
\(
v_{x}=\int d v=\int_{t=0}^{T}\left[f_{0}\left(1-\frac{t}{T}\right)\right] d t
\)

\(\begin{aligned}
&=f_{0}\left[\left(t-\frac{t^{2}}{2 T}\right)\right]_{0}^{T} \\
&=f_{0}\left(T-\frac{T^{2}}{2 T}\right)=f_{0}\left(T-\frac{T}{2}\right)=\frac{1}{2} f_{0} T
\end{aligned}\)

Hint

At \(\mathrm{t}=0\), particle is at, let’s say \(\mathrm{x}\) distance, from \(\mathrm{O}\); then putting \(\mathrm{t}=0\) in the given displacement-time equation we get; \(x=40+12(0)-(0)^3=40 \mathrm{~m}\)
Particle comes to rest that means velocity of particle becomes zero after travelling certain displacement; let’s say the time be t. then after differentiating the given displacement-time equation wrt. time we get velocity-time equation
\(
\mathrm{v}=12-3 \mathrm{t}^2
\)
at time \(\mathrm{t}=\mathrm{t}\) (the time when the particle comes to rest) :
\(
\begin{aligned}
&\mathrm{v}=0 ; \\
&=>12-3 \mathrm{t}^2=0 ; \\
&=>\mathrm{t}=2 \mathrm{~s}
\end{aligned}
\)
Then ,at \(\mathrm{t}=2 \mathrm{~s}\) we are at, let’s say \(\mathrm{x}^{\prime}\) distance from \(\mathrm{O}\);
put this value of t \((=2)\) in given displacement-time equation,
we get;
\(
\begin{aligned}
&\mathrm{x}^{\prime}=40+12(2)-(2)^3 \\
&=56 \mathrm{~m}
\end{aligned}
\)
Further;
We have seen that the particle started his journey when it is at \(40 \mathrm{~m}\) from the point \(\mathrm{O}\).
And came to rest at \(56 \mathrm{~m}\) from the point \(\mathrm{O}\).
then the particle traveled a distance of:
\(
56-40=16 \mathrm{~m} .
\)

Hint

Given, \(x=a e^{-\alpha t}+b e^{\beta t}\)
Velocity \(v=\frac{d x}{d t}=-a \alpha e^{-\alpha t}+b \beta e^{\beta t}\)
\(
=A+B
\)
where, \(A=-a \alpha e^{-\alpha t}\)
\(
B=b \beta e^{\beta t}
\)
The value of term \(A=-a \alpha e^{-\alpha t}\) decreases and of term \(B=b \beta e^{\text {pt }}\) increases with time. As a result, velocity goes on increasing with time.

Hint

Displacement
\(
s=3 t^{3}+7 t^{2}+5 t+8
\)
Velocity \(=\frac{d s}{d t}=9 t^{2}+14 t+5\)
Acceleration \(=\frac{d^{2} s}{d t^{2}}=18 t+14\)
Acceleration at \((t=1 \mathrm{~s})\)
\(
=18 \times 1+14=18+14=32 \mathrm{~m} / \mathrm{s}^{2}
\)

Hint

According to the conservation of energy, the kinetic energy of a car = work done in stopping the car
i.e. \(\frac{1}{2} m v^{2}=F s\)
where \(F\) is the retarding force and \(s\) is the stopping distance.
For the same retarding force,

\(\begin{aligned}
s & \propto v^{2} \\
\frac{s_{2}}{s_{1}} &=\left(\frac{v_{2}}{v_{1}}\right)^{2}=\left(\frac{80}{40}\right)^{2}=4 \\
s_{2} &=4 s_{1}=4 \times 2 \\
&=8 \mathrm{~m}
\end{aligned}\)

Hint

\(\text { distance, } s=a t^{2}-b t^{3}\)

velocity, \(v=\frac{d s}{d t}=2 a t-3 b t^{2}\) acceleration \(a=\frac{d v}{d t}=2 a-6 b t\) Acceleration is zero at
\(
2 a-6 b t=0 \quad \Rightarrow \quad t=\frac{a}{3 b}
\)

Hint

In the Figure above
\(
\mathrm{AA}_{1}=v_{\text {max. }}=\alpha t_{1}=\beta \mathrm{t}_{2}
\)
But \(t=t_{1}+t_{2}=\frac{v_{\max }}{\alpha}+\frac{v_{\max }}{\beta}\)
\(
=v_{\max }\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=v_{\max }\left(\frac{\alpha+\beta}{\alpha \beta}\right)
\)
or, \(v_{\max }=t\left(\frac{\alpha \beta}{\alpha+\beta}\right)\)

Hint

At \(E\), the slope of the curve is negative.
The line bending towards the time axis represents decreasing velocity or negative velocity of the particle.

Hint

Velocity, \(v=\frac{d s}{d t}=3 t^{2}-12 t+3\)
Acceleration, \(a=\frac{d v}{d t}=6 t-12 ;\) For \(a=0\), we have, \(0=6 t-12\) or \(t=2 \mathrm{~s}\). Hence, at \(t=2\) s the velocity will be
\(
v=3 \times 2^{2}-12 \times 2+3=-9 \mathrm{~ms}^{-1}
\)

Hint

\(\frac{\mathrm{S}_{4}}{\mathrm{~S}_{3}}=\frac{0+\frac{a}{2}(2 \times 4-1)}{0+\frac{a}{2}(2 \times 3-1)}=\frac{7}{5}\)
Equation of the motion of uniformly accelerated motion, the distance travelled in \(n^{\text {th }} \sec\) is given by,
\(
S_{n}=u+\frac{a}{2}(2 n-1)
\)

Hint

In one dimensional motion, the body can have at a time one velocity but not two values of velocities. Figure with the circle is the right choice.

Hint

Let \(x\) be the total distance between points \(P\) and \(Q\) and \(v\) be the velocity of car while passing a certain middle point of PQ. If \(a\) is the acceleration of the car, then
For part \(P Q\),
\(
\begin{gathered}
40^{2}-30^{2}=2 a x \\
a=\frac{350}{x}
\end{gathered}
\)
For part RQ,
\(
40^{2}-v^{2}=\frac{2 a x}{2}
\)
Putting value of \(a\) from the above analysis we have
\(
\begin{aligned}
40^{2}-v^{2} &=2\left(\frac{350}{x}\right) \frac{x}{2} \\
\text { or } \quad 40^{2}-v^{2} &=350 \text { or } \\
v^{2} &=1250 \\
\Rightarrow \quad & v =25 \sqrt{2} \mathrm{~km} / \mathrm{h}
\end{aligned}
\)

Hint

Let \(v\) be the relative velocity of scooter w.r.t bus as \(v=v_{S} – v_{B}\)
\(
\begin{aligned}
&v=\frac{1000}{100}=10 \mathrm{~ms}^{-1}, \mathrm{v}_{\mathrm{B}}=10 \mathrm{~ms}^{-1} \\
&\begin{aligned}
\therefore & v_{\mathrm{S}}=v+v_{\mathrm{B}} \\
& \quad = 10+10 =20 \mathrm{~ms}^{-1} \\
& \quad \text { velocity of scooter }=20 \mathrm{~ms}^{-1}
\end{aligned}
\end{aligned}
\)

Hint

Relative velocity of parrot w.r.t the train \(=10-(-5)=15 \mathrm{~ms}^{-1}\).
Time taken by parrot to cross the train \(=\frac{150}{15}=10 \mathrm{~s}\)
When two bodies are moving in a direction relative velocity is the sum of individual velocities.
\(
\vec{v}_{A B}=\vec{v}_{A}-\vec{v}_{B}
\)
The distance that a bird has to travel to cross the train is equal to the length of the train. Now since the train and the bird are travelling in the opposite direction, therefore the speed will be added up.

Hint

Given, \(u=20 \mathrm{~m} / \mathrm{s}, \mathrm{v}=80 \mathrm{~m} / \mathrm{s}\) and \(h=\) ?
From kinematic equation of motion,
\(
\begin{aligned}
& v^{2}=u^{2}+2 g h \\
\Rightarrow h &=\frac{v^{2}-u^{2}}{2 g} \\
&=\frac{(80)^{2}-(20)^{2}}{2 \times 10}\left(\because \text { given, } g=10 \mathrm{~m} / \mathrm{s}^{2}\right) \\
&=300 \mathrm{~m}
\end{aligned}
\)

Hint

\(\begin{array}{ll}
\text { (a) } & \because \quad h=\frac{1}{2} g t^{2} \\
\therefore & h_{1}=\frac{1}{2} g(5)^{2}=125 \\
& h_{1}+h_{2}=\frac{1}{2} g(10)^{2}=500 \\
\Rightarrow & h_{2}=375 \\
& h_{1}+h_{2}+h_{3}=\frac{1}{2} g(15)^{2}=1125 \\
\Rightarrow & h_{3}=625 \\
& h_{2}=3 h_{1}, h_{3}=5 h_{1} \\
\text { or } & h_{1}=\frac{h_{2}}{3}=\frac{h_{3}}{5}
\end{array}\)

Hint

Given, \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) and \(h=20 \mathrm{~m}\)
We have \(v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 20}\)
\(
=\sqrt{400}=20 \mathrm{~m} / \mathrm{s}
\)

Hint

For the first ball, \(u=0\)
\(
\therefore \quad s_{1}=\frac{1}{2} g t_{1}^{2}=\frac{1}{2} \times g(18)^{2}
\)
For second ball, initial velocity \(=v\)
\(\therefore \quad s_{2}=v t_{2}+\frac{1}{2} g t^{2}\)
\(t_{2}=18-6=12 \mathrm{~s}\)
\(\Rightarrow \quad s_{2}=v \times 12+\frac{1}{2} g(12)^{2}\)
\(
\begin{aligned}
& \text { Here, } \quad s_{1} =s_{2} \\
\frac{1}{2} g(18)^{2} &=12 v+\frac{1}{2} g(12)^{2} \\
\Rightarrow \quad v &=75 \mathrm{~ms}^{-1}
\end{aligned}
\)

Hint

No external force is acting, therefore, momentum is conserved. By momentum conservation, \(50 u+0.5 \times 2=0\)
where \(u\) is the velocity of man.
\(
u=-\frac{1}{50} \mathrm{~ms}^{-1}
\)
Negative sign of \(u\) shows that man moves upward.
Time taken by the stone to reach the ground \(=\frac{10}{2}=5 \mathrm{sec}\)

Distance moved by the man \(=5 \times \frac{1}{50}=0.1 \mathrm{~m}\) \(\therefore \quad\) when the stone reaches the floor, the distance of the man above floor \(=10.1 \mathrm{~m}\)

Hint

Let \(t_{1} \& t_{2}\) be the time taken by \(A\) and \(B\) respectively to reach the ground then from the formula,
\(
h=\frac{1}{2} g t^{2}
\)
For first body, \(\quad 16=\frac{1}{2} g t_{1}^{2}\)
For second body, \(25=\frac{1}{2} g t_{2}^{2}\)
\(
\therefore \frac{16}{25}=\frac{t_{1}^{2}}{t_{2}^{2}} \Rightarrow \frac{t_{1}}{t_{2}}=\frac{4}{5}
\)
If a body is dropped from some height, the motion is independent of the mass of the body. The time taken to reach the ground \(t=\sqrt{2 h / g}\) and final velocity \(V=\sqrt{2 g h}\) and initial velocity, \(u=0\).

Hint

For part \(A B\)
From 3rd equation of motion \(v^{2}=u^{2}-2 \mathrm{gH}\)\(\begin{aligned}
&0=u^{2}-2 \mathrm{~g}(\mathrm{H} / 2)=u^{2}-\mathrm{gH} \\
&\mathrm{H}=\frac{u^{2}}{g}=\frac{10^{2}}{10}=10 \mathrm{~m}
\end{aligned}\)

Hint

Let body takes T sec to reach maximum height. Then \(v=u-g T\)
\(v=0\), at highest point.
So, \(T=\frac{u}{g}\)
Velocity attained by body
in \((T-t) \sec v=u-g(T-t)\)
\(v=u-g \mathrm{~T}+g t=u-g \frac{u}{g}+g t \quad(\because \mathrm{T}=u / g)\)
or \(v=g t\)
\(\therefore\) Distance travelled in last \(t \sec\) of its ascent
\(s=v t-\frac{1}{2} g t^{2}\)
\(
s=(g t) t-\frac{1}{2} g t^{2}=\frac{1}{2} g t^{2}
\)

If a body is projected vertically upward, then, final velocity, \(v=0\); intial speed, \(u=g t\) and height attained, \(h=\frac{u^{2}}{2 g}=\frac{g^{2} t^{2}}{2 g}=\frac{1}{2} g t^{2}\)

Hint

From equation of motion time taken by ball to reach maximum height \(v=u-g t\)
At maximum height,
final speed is zero i.e. \(v=0\)
So, \(u=g t\) or \(t=\frac{u}{g}\)
In \(2 \mathrm{~s}, u=2 \times 9.8=19.6 \mathrm{~m} / \mathrm{s}\)
If man throws the ball with velocity of \(19.6 \mathrm{~m} / \mathrm{s}\) then after \(2 \mathrm{~s}\) it will reach the maximum height. When he throws 2 nd ball, Ist is at the top. When he throws the third ball, 1st will come to the ground and 2nd will be at the top. Therefore, only 2 balls are in the air. If he wants to keep more than 2 balls in the air he should throw the ball with a speed greater than \(19.6 \mathrm{~m} / \mathrm{s}\).

Hint

Initial velocity \((u)=40 \mathrm{~m} / \mathrm{s}\)
Acceleration \(a=-g \mathrm{~m} / \mathrm{s}^{2}=-10 \mathrm{~m} / \mathrm{s}^{2}\)
Time \(=2\) seconds
By first equation of the motion,
\(
\begin{aligned}
&v=u+a t \\
&v=40-10(2)=20 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

Time taken by the body to reach the ground from some height is the same as taken to reach that height. Hence, the time to reach the ground from its maximum height is \(5 \mathrm{~s}\).

Hint

Initial velocity \((u)=0\); Time \((t)=4\) sec and gravitational acceleration \((g)=10 \mathrm{~m} / \mathrm{s}^{2}\).
Height of tower
\(
h=u t+\frac{1}{2} g t^{2}=(0 \times 4)+\frac{1}{2} \times 10 \times(4)^{2} .=80 \mathrm{~m} \text {. }
\)

Hint

The speed of an object, falling freely due to gravity, depends only on its height and not on its mass \(V=\sqrt{2 g h}\). Since the paths are frictionless and all the objects fall through the same height, therefore, their speeds on reaching the ground will be in the ratio of \(1: 1: 1\)

Hint

Let \(t\) be the time interval of two drops.
For the third drop to fall
\(5=\frac{1}{2} g(2 t)^{2}\)
\([A s  u=0\) ]
or \(\frac{1}{2} g t^{2}=\frac{5}{4}\)
Let \(x\) be the distance through which second drop falls for time t, then
\(
x=\frac{1}{2} g t^{2}=\frac{5}{4} m
\)
Thus, height of second drop from ground
\(
=5-\frac{5}{4}=\frac{15}{4}=3.75 \mathrm{~m}
\)

Hint

Let the body fall through the height of tower in \(n\)th seconds. From,
\(D_{n}=u+\frac{a}{2}(2 n-1)\) we have, total distance travelled in last 2 seconds of fall is
\(
\begin{aligned}
&D=D_{t}+D_{(t-1)} \\
&=\left[0+\frac{g}{2}(2 n-1)\right]+\left[0+\frac{g}{2}\{2(n-1)-1\}\right] \\
&=\frac{g}{2}(2 n-1)+\frac{g}{2}(2 n-3)=\frac{g}{2}(4 n-4) \\
&=\frac{10}{2} \times 4(n-1)
\end{aligned}
\)
or, \(40=20(n-1)\) or \(n=2+1=3 \mathrm{~s}\)
Distance travelled in \(t\) seconds is
where, \(t=3 \mathrm{sec}\)
\(
s=u t+\frac{1}{2} a t^{2}=0+\frac{1}{2} \times 10 \times 3^{2}=45 \mathrm{~m}
\)

Hint

\(\frac{x(4)}{x(5)}=\frac{\frac{g}{2}(2 \times 4-1)}{\frac{g}{2}(2 \times 5-1)}=\frac{7}{9}\)
\(
\left[\because S_{n} \mathrm{th}=u+\frac{a}{2}(2 n-1) \text { and } u=0, a=g\right]
\)
In the absence of air resistance, all bodies irrespective of the size, weight, or composition fall with the same acceleration near the surface of the earth. This motion of a body falling towards the earth from a small altitude \((\mathrm{h}<<\mathrm{R})\) is called free fall.

Distance travelled in nth second \(h_{n}=\frac{g}{2}(2 n-1)\)

Hint

Given, the initial velocity of a car, \(u=0\)
The acceleration of a car, \(a=5 \mathrm{~m} / \mathrm{s}^{2}\)
Thus, the final velocity of car at \(t=4 \mathrm{~s}\) is \(20 \mathrm{~m} / \mathrm{s}\).
At \(t=4 s_{\text {, the ball is dropped out of a }}\) window by a person sitting in the car.
The velocity of the ball in the \(x\)-direction,
\(v_{x}=20 \mathrm{~m} / \mathrm{s}\) (due to the car)
Therefore, in the \(y\)-direction, the
acceleration is equal to the acceleration due to gravity,
\(
a_{y}=g=10 \mathrm{~m} / \mathrm{s}^{2}
\)
The velocity of the ball in the \(y\)-direction,
\(
\begin{aligned}
\quad v_{y} &=u+a_{y} t \quad \Rightarrow v_{y}=0+10 \times 2 \\
\Rightarrow v_{y} &=20 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Thus, the velocity of the ball in \(y\)-direction is \(20 \mathrm{~m} / \mathrm{s}\).
The net velocity at \(t=6 \mathrm{~s}\),
\(
\begin{aligned}
\quad v &=\sqrt{v_{x}^{2}+v_{y}^{2}} \Rightarrow v=\sqrt{(20)^{2}+(20)^{2}} \\
\Rightarrow \quad v &=20 \sqrt{2} \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Thus, the velocity of the ball at \(t=6 \mathrm{~s}\) is \(20 \sqrt{2} \mathrm{~m} / \mathrm{s}\) and there is no acceleration in the \(x\)-direction, \(a_{x}=0 \mathrm{~ms}^{-2}\)
In \(y\)-direction, \(a_{y}=10 \mathrm{~ms}^{-2}\)
Now, we shall determine the net acceleration \(\Rightarrow a=6 \mathrm{~s}, \quad a=\sqrt{a_{x}^{2}+a_{y}^{2}}\) \(\Rightarrow a=\sqrt{(0)+(10)^{2}} \Rightarrow a=10 \mathrm{~ms}^{-2}\)

Hint

Distance covered nth seconds is \(s_{n}\). Distance covers in \((n+1)\) th seconds is \(s_{n+1}\).
Initial velocity of small block, \(u=0\)
Distance cover in nth seconds,
\(
\begin{aligned}
s_{n} &=u+\frac{a}{2}(2 n-1) \\
\Rightarrow s_{n} &=0+\frac{a}{2}(2 n-1) \\
\Rightarrow s_{n} &=\frac{a}{2}(2 n-1)
\end{aligned}
\)
Distance cover in \((n+1)\) th seconds,
\(
\begin{aligned}
s_{n+1} &=u+\frac{a}{2}[2(n+1)-1] \\
\Rightarrow s_{n+1} &=0+\frac{a}{2}(2 n+2-1) \\
\Rightarrow s_{n+1} &=\frac{a}{2}(2 n+1)
\end{aligned}
\)
On dividing \(s_n  with  s_{n+1}\)
\(
\begin{gathered}
\frac{s_{n}}{s_{n+1}}=\frac{\frac{a}{2}(2 n-1)}{\frac{a}{2}(2 n+1)} \\
\Rightarrow \quad \frac{s_{n}}{s_{n+1}}=\frac{(2 n-1)}{(2 n+1)}
\end{gathered}
\)

Hint

Slope of \(x\)-t curves gives the velocity
\(
\Rightarrow \text { Ratio }=\frac{\tan 30^{\circ}}{\tan 45^{\circ}}=\frac{1}{\frac{\sqrt{3}}{1}}=1: \sqrt{3}
\)

Hint

\(
\begin{aligned}
v_{\text {avg }} & =\frac{2 v_1 v_2}{v_1+v_2} \\
& =\frac{2 \times v \times 2 v}{v+2 v} \\
& =\frac{4 v}{3}
\end{aligned}
\)