Motion in a Straight Line

No option is matching. displacement of car in interval, $0 \leq t^{\prime} \leq t$ is $s=u t+\frac{1}{2} a t^2$

Here, $u=10 \mathrm{~m} / \mathrm{s}$ and $a=2 \mathrm{~m} / \mathrm{s}^2$

Now,

Using equation, $s=u t+\frac{1}{2} a t^2$

Displacement of car in 3 s

$ =s_3=30+\frac{1}{2} \times 2 \times(3)^2=39 \mathrm{~m} $

Displacement in 4 s

$ =s_4=40+\frac{1}{2} \times 2 \times 4^2=56 \mathrm{~m} $

Displacement in 5 s

$ =s_5=50+\frac{1}{2} \times a \times 5^2=75 \mathrm{~m} $

So, distance $(3 \leq t \leq 4)=s_1=s_4-s_3$

$ =56-39=17 \mathrm{~m} $

And distance $(4 \leq t \leq 5)=s_2=s_5-s_4$

$ =75-56=19 \mathrm{~m} $

Hence, ratio $\frac{s_2}{s_1}=\frac{19 \mathrm{~m}}{17 \mathrm{~m}}=\frac{19}{17}$

Ball has same displacement $H$ at $t=2 \mathrm{~s}$ and $t=10 \mathrm{~s}$

Now we use,

$ s=u t+\frac{1}{2} a t^2 $

For $t=2 \mathrm{~s}$, (upward motion of ball)

$ \begin{aligned} & & H & =u \times 2+\frac{1}{2}(-9.8) \times 2^2 \\ \Rightarrow & & H & =2 u-19.6 \end{aligned} $

For $\quad t=10 \mathrm{~s}$

(downward motion of ball)

$ \begin{aligned} & & H & =u \times 10+\frac{1}{2}(-9.8) \times 10^2 \\ \Rightarrow \quad & & H & =10 u-490 \end{aligned} $

From Eqs. (i) and (ii), we have

$ \begin{aligned} & & 10 u-490 & =2 u-19.6 \\ \Rightarrow & & 8 u & =470.4 \\ \Rightarrow & & u & =58.8 \mathrm{~m} / \mathrm{s} \end{aligned} $

Substituting $u$ in Eq. (i), we get

$ \begin{aligned} & H=2 \times(58.8)-19.6 \\ \Rightarrow \quad & H=98 \mathrm{~m} \end{aligned} $

The given situation is shown below.

Now, speed of bus in $X$ to $Y$-direction observed by $\operatorname{man}=v-u$ and speed of bus in $Y$ to $X$-direction observed by man $=v+u$ Buses moves in either direction with equal speed $v$ and at equal time intervals $T$. So, distance between buses coming from either direction is same.

So, $\quad$ speed × time $=$ speed × time $(X$ to $Y)(Y$ to $X)$

$ \begin{aligned} \Rightarrow &\,\,\,\,\,\,\,\,\,\,\,\,\,\, (v-u) t_1 =(v+u) t_2 \\ \Rightarrow &\,\,\,\,\,\,\,\,\,\,\,\,\,\, v\left(t_1-t_2\right) =u\left(t_1+t_2\right) \\ \text { or } &\,\,\,\,\,\,\,\,\,\,\,\,\,\, u =\frac{v\left(t_1-t_2\right)}{t_1+t_2} \end{aligned} $

Hence, distance $D$ between any 2-buses is

$ \begin{aligned} D & =\left(v-i_1\right) t_1 \\ & =\left(v-\frac{v\left(t_1-t_2\right)}{t_1+t_2}\right) t_1 \\ & =v\left(\frac{t_1+t_2-t_1+t_2}{t_1+t_2}\right) t_1 \\ D & =v\left(\frac{2 t_1 t_2}{t_1+t_2}\right) \end{aligned} $

Hence, time interval between any 2 -buses is

$ T=\frac{\text { Distance }}{\text { Speed }}=\frac{D}{v}=\frac{2 t_1 t_2}{t_1+t_2} $

Initial velocity of rocket, $u=0 \mathrm{~m} / \mathrm{s}$ acceleration, $a=20 \mathrm{~m} / \mathrm{s}^2$

Upward velocity of rocket at the end 5 s ,

$ \begin{aligned} v & =u+a t \\ & =0+20 \times 5=100 \mathrm{~m} / \mathrm{s} \end{aligned} $

Distance travelled by rocket in 5 s ,

$ s_1=\frac{1}{2} a t^2=\frac{1}{2} \times 20 \times 5^2=250 \mathrm{~m} $

After 5 s , rocket moves under the effect of gravitational acceleration and with initial velocity of $100 \mathrm{~m} / \mathrm{s}$.

∴ Distance travelled by rocket in this situation is $s_2$, then using equation,

$ \begin{array}{rlrl} & v^2 =u^2-2 g s_2 \\ \Rightarrow & 0 =100^2-2 g s_2 \\ \Rightarrow & s_2 =\frac{100^2}{2 g}=\frac{100 \times 100}{2 \times 10}=500 \mathrm{~m} \end{array} $

∴ Total upward distance covered by rocket,

$ s=s_1+s_2=250+500=750 \mathrm{~m} $

Now, rocket starts falling from this height and if $v^{\prime}$ be the velocity of rocket when it reaches at ground, then

$ \begin{aligned} \left(v^{\prime}\right)^2 & =u^2+2 g s \\ & =0+2 \times 10 \times 750=15000 \\ \Rightarrow \quad v^{\prime} & =\sqrt{15000} \\ & =50 \sqrt{6} \mathrm{~m} / \mathrm{s} \end{aligned} $

Displacement of the particle,

$ \text { } \begin{aligned} x & =\alpha t^3+\beta t^2+\gamma \\at\,\,\,\,\,\,\, t_1=1 \mathrm{~s}, x_1 & =\alpha \cdot 1^3+\beta \cdot 1^2+\gamma \\ & =\alpha+\beta+\gamma \\ \text { at } \quad t_2=3 \mathrm{~s}, x_3 & =\alpha \cdot 3^3+\beta \cdot 3^2+\gamma \\ & =27 \alpha+9 \beta+\gamma \end{aligned} $

Average velocity,

$ \begin{aligned} v_1 & =\frac{\text { Total distance travelled }}{\text { Total time taken }} \\ & =\frac{x_3-x_1}{t_2-t_1} \\ & =\frac{27 \alpha+9 \beta+\gamma-\alpha-\beta-\gamma}{3-1} \\ & =\frac{26 \alpha+8 \beta}{2}=13 \alpha+4 \beta \end{aligned} $

Instantaneous velocity at $t=3 \mathrm{~s}$ is given as,

$ \begin{aligned} v_2=\left.\frac{d x}{d t}\right|_{t=3 s} & =\left(3 \alpha t^2+2 \beta t+0\right) / \mathrm{at} t=3 \mathrm{~s} \\ & =3 \alpha(3)^2+2 \beta(3)=27 \alpha+6 \beta \\ \therefore \quad \frac{v_1}{v_2} & =\frac{13 \alpha+4 \beta}{27 \alpha+6 \beta} \end{aligned} $

$ \text { The given situation is shown below } $

$O=$ position of observer

$C=$ initial position of aeroplane at $t=0$

$A=$ final position of aeroplane after time $t$.

∴ Distance travelled by aeroplane after time $t$,

$d=C A=O B=$ Velocity × Time $=v t$

In $\triangle A O B$,

$ \tan \theta=\frac{O B}{A B}=\frac{v t}{h} \Rightarrow \theta=\tan ^{-1}(v t / h) $

We know that, area under $a-t$ graph gives us change in velocity

Given; $u=0 \mathrm{~m} / \mathrm{s}$

∴ Area under $a$ - $t$ graph $=$ Change in velocity

$ \begin{aligned} \Rightarrow & & \frac{1}{2} \times 15 \times 10 & =v-u \\ \Rightarrow & & 75 & =v-0 \\ \Rightarrow & & v & =75 \mathrm{~m} / \mathrm{s} \end{aligned} $

Assertion is false because if we take the example of a ball thrown vertically upwards, then at the topmost point of its motion, the ball momentarily comes to rest but acceleration due to gravity keeps on acting on the ball.

Reason is true because whenever any object reverses its direction of motion, then the sign of velocity changes because it comes to rest for a moment at the point where it changes its direction of motion. The same example of motion of ball given above can be used to explain it as the ball changes the direction of motion after reaching the topmost point when its speed becomes zero for a moment.

Speed of river $=v$

Speed of man in still water $=2 v$

∴ Speed upstream $=(2 v-v)$

Speed downstream $=(2 v+v)$

Now, according to the question, total time taken in going upstream and back downstream to the starting point $=t$

$ \therefore \quad t_{\text {up }}+t_{\text {down }}=t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Let the total distance on one way $=d$

$ \therefore \quad t_{\text {up }}=\frac{d}{2 v-v} \Rightarrow t_{\text {down }}=\frac{d}{2 v+v} $

Putting these values is Eq. (i) we get,

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{d}{2 v-v}+\frac{d}{2 v+v}=t $

$ \begin{array}{ll} \Rightarrow & d\left[\frac{1}{v}+\frac{1}{3 v}\right]=t \Rightarrow \frac{d}{v}\left[\frac{3+1}{3}\right]=t \\ \Rightarrow & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=\frac{4 d}{3 v} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

It is given that time taken by the man to cover the same distance in still water

$ \Rightarrow \quad t_0=\frac{2 d}{2 v}=\frac{d}{v} $

∴ From Eq. (ii) $t=\frac{4}{3} t_0 \Rightarrow \frac{t}{t_0}=\frac{4}{3}$

Initial velocity,

$ u=0 \mathrm{~m} / \mathrm{s}\,\,\,\, [ ∵ rest]$

Final velocity, $v=10 \mathrm{~m} / \mathrm{s}$

Time taken, $t=2 \mathrm{~s}$

We know,

$ a=\frac{v-u}{t}=\frac{10-0}{2}=5 \mathrm{~m} / \mathrm{s}^2 $

Now, $s=u t+\frac{1}{2} a t^2$

$ =0+\frac{1}{2} \times 5 \times 4=10 \mathrm{~m} $

$ \text { Let } u \text { be initial velocity of bullet, } $

Final velocity of bullet after travelling a distance $x$ is equal to $\left(u-\frac{u}{3}\right)=\frac{2 u}{3}$

We know, $v^2=u^2+2 a s$

$ \begin{aligned} & \Rightarrow \quad\left(\frac{2 u}{3}\right)^2=u^2+2 a x \\ & \Rightarrow \quad \frac{4 u^2}{9}-u^2=2 a x \\ & \Rightarrow \quad x=\frac{-5}{18 a} u^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Again after embedding further distance of $x^{\prime}$, it comes to rest

$ v^{\prime 2}=v^2+2 a x^{\prime} $

(Now, $v=$ initial velocity for $x^{\prime}$ distance)

$ \begin{array}{ll} \Rightarrow & 0=\frac{4 u^2}{9}+2 \times a \times x^{\prime} \\ \Rightarrow & x^{\prime}=\frac{-4 u^2}{18 a} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \Rightarrow & \frac{x^{\prime}}{x}=\frac{-4 u^2}{\frac{18 a}{-5 u^2}}=\frac{4}{5} \end{array} $

[from Eqs. (i) and (ii)]

Let $v$ be the minimum velocity of student so, that he could catch the bus

If student catch the bus in time $t$, then distance travelled by student in time $t =16+$ distance travelled by bus in time $t$.

$\Rightarrow v t=16+\left(u t+\frac{1}{2} a t^2\right)$

$\Rightarrow v t=16+0 \times t+\frac{1}{2} \times 9 \times t^2$

$\Rightarrow v t=16+\frac{9}{2} t^2 \Rightarrow 9 t^2-2 v t+32=0$

The above equation must have real roots. i.e its discriminant $\geq 0$

$\begin{aligned} & \text { i.e }(2 v)^2-4 \times 9 \times 32 \geq 0 \\ & \Rightarrow 4 v^2-4 \times 288 \geq 0 \Rightarrow v^2-288 \geq 0 \\ & \Rightarrow v^2 \geq 288 \\ & \quad v \geq 12 \sqrt{2} \mathrm{~m} / \mathrm{s} \end{aligned}$

Minimum velocity of student to catch the bus $=12 \sqrt{2} \mathrm{~m} / \mathrm{s}=\alpha \sqrt{2} \mathrm{~m} / \mathrm{s}$ (given)

$\therefore \alpha=12$

The given situation is shown below

At time $t=0$ object is at point $A$ with a distance of 3 cm.

At time $t=2 \mathrm{~s}$ object reaches from point $A$ to point $B$. i.e in negative direction of $X$-axis.

$\therefore$ Distance travelled by the object in 2 s,

$\begin{aligned} d & =3-(-5)=3+5=8 \mathrm{~cm} \\ d & =8 \hat{i} \end{aligned}$

Initial velocity of object along $X$-axis,

$\mathbf{v}=-12 \hat{i} \mathrm{~cm} / \mathrm{s}$

By using second equation of motion,

$\begin{aligned} \mathbf{d} & =\mathbf{u} t+\frac{1}{2} \mathbf{a} t^2 \\ \Rightarrow 8 \hat{i} & =-12 \hat{i} \times 2-\frac{1}{2} \mathbf{a} \times 4 \Rightarrow 8 \hat{i}=-24 \hat{i}-2 \mathbf{a} \\ \Rightarrow 2 \mathrm{a} & =-32 \hat{i} \Rightarrow \mathbf{a}=-16 \hat{i} \mathrm{~cm} / \mathrm{s}^2 \end{aligned}$

Displacement equation of ball moving in vertical plane, $y=P t^2-Q t^3$

The ball will reach at maximum height, if

$\frac{d y}{d t}=0$

$\begin{aligned} & & \frac{d}{d t}\left(P t^2-Q t^3\right) & =0 \\ \Rightarrow & & 2 P t-3 Q t^2 & =0 \Rightarrow t(2 P-3 Q t)=0 \\ \therefore & & t \neq 0,2 P-3 Q t & =0 \\ \Rightarrow & & t & =\frac{2 P}{3 Q} \end{aligned}$

$\because$ Maximum height attained by the ball,

$\begin{aligned} y_{\max }=P t^2 & -Q t^3 \\ \text { When, } t & =\frac{2 P}{3 Q} \\ & =P\left(\frac{2 P}{3 Q}\right)^2-Q\left(\frac{2 P}{3 Q}\right)^3 \\ & =\frac{4 P^3}{9 Q^2}-\frac{8 P^3}{27 Q^2}=\frac{4 P^3}{27 Q^2} \end{aligned}$

Suppose the car covers distance $x \mathrm{~km}$ from location $A$ to $B$ with speed of $60 \mathrm{~kmh}^{-1}$.

$\begin{array}{ll} \text { i.e } & v_{A B}=60 \mathrm{~km} / \mathrm{h} \\ \text { and } & v_{B A}=v \mathrm{~km} / \mathrm{h} \end{array}$

$\begin{aligned} \text { Average speed } & =\frac{\text { total distance }}{\text { total time }} \\ 48 & =\frac{x+x}{t_{A B}+t_{B A}} \quad \text{... (i)} \end{aligned}$

Now, $\quad t_{A B}=\frac{x}{60}$ and $t_{B A}=\frac{x}{v}$

From Eq.(i), we have

$\begin{aligned} 48 & =\frac{x+x}{\frac{x}{60}+\frac{x}{v}} \\ \Rightarrow \quad 48 & =\frac{2}{\frac{1}{60}+\frac{1}{v}} \Rightarrow 48=\frac{2 \times 60 v}{v+60} \\ \Rightarrow 48 v+2880 & =120 v \Rightarrow 72 v=2880 \\ v & =\frac{2880}{72} \\ & =40 \mathrm{~km~h}^{-1} \end{aligned}$

The relationship between time $ t $ and distance $ x $ for a moving body is given by $ t = mx^2 + nx $, where $ m $ and $ n $ are constants. To determine the retardation (negative acceleration) of the motion, let's follow the steps to derive it:

Given:

$ t = mx^2 + nx $

First, differentiate $ t $ with respect to $ x $:

$ \frac{dt}{dx} = 2mx + n $

Since velocity $ v $ is defined as $ \frac{dx}{dt} $, we can write:

$ \frac{1}{v} = \frac{dt}{dx} = 2mx + n $

Thus,

$ v = \frac{1}{2mx + n} $

Next, to find the acceleration $ a $ (which is the derivative of velocity with respect to time), we start with the chain rule:

$ \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} $

Since $ v = \frac{dx}{dt} $, substituting $ x $ gives:

$ \frac{dx}{dt} = v $

Rewrite it:

$ \frac{dv}{dt} = \frac{dv}{dx} \cdot v $

Differentiate $ v $ with respect to $ x $:

$ v = (2mx + n)^{-1} $

$ \frac{dv}{dx} = -\frac{2m}{(2mx + n)^2} $

Then,

$ \frac{dv}{dt} = -\frac{2m}{(2mx + n)^2} \cdot v $

Substitute $ v = \frac{1}{2mx + n} $ into the expression:

$ \frac{dv}{dt} = -\frac{2m}{(2mx + n)^2} \cdot \frac{1}{2mx + n} $

Simplify the expression:

$ \frac{dv}{dt} = -2m \left( \frac{1}{2mx + n} \right)^3 $

$ a = -2m v^3 $

So, the retardation (negative acceleration) is:

$ a = -2m v^3 $

The slope of the given v versus x graph is $m = - {{{v_0}} \over {{x_0}}}$ and intercept is c = + v₀. Hence, v varies with x as

$v = - \left( {{{{v_0}} \over {{x_0}}}} \right)x + {v_0}$ ...... (1)

where v₀ and x₀ are constants of motion. Differentiating with respect to time t, we have

${{dv} \over {dt}} = - \left( {{{{v_0}} \over {{x_0}}}} \right){{dx} \over {dt}}$

or $a = - \left( {{{{v_0}} \over {{x_0}}}} \right)v$ ........ (2)

Using Eq. (1) in Eq. (2), we get

$a = - \left( {{{{v_0}} \over {{x_0}}}} \right)\left( { - {{{v_0}} \over {{x_0}}}x + {v_0}} \right)$

$a = {\left( {{{{v_0}} \over {{x_0}}}} \right)^2}x - {{v_0^2} \over {{x_0}}}$

Thus the graph of a versus x is a straight line having a positive slope = ${\left( {{{{v_0}} \over {{x_0}}}} \right)^2}$ and negative intercept = $ - {{v_0^2} \over {{x_0}}}$. Hence the correct choice is (a).

The velocity of a particle is given by $ v = v_0 + gt + Ft^2 $. Its position is $ x = 0 $ at $ t = 0 $. To find its displacement after time $ t = 1 $, follow these steps:

Given:

$ v = v_0 + gt + Ft^2 $

We know that:

$ \frac{dx}{dt} = v_0 + gt + Ft^2 $

To find the displacement, integrate both sides with respect to $ t $:

$ \int_{x = 0}^{x} dx = \int_{t = 0}^{t = 1} (v_0 + gt + Ft^2) \, dt $

This simplifies to:

$ x = \left[ v_0 t + \frac{gt^2}{2} + \frac{Ft^3}{3} \right]_{t = 0}^{t = 1} $

Evaluating the integral from $ t = 0 $ to $ t = 1 $:

$ x = v_0 + \frac{g}{2} + \frac{F}{3} $

Therefore, the displacement after time $ t = 1 $ is:

$ x = v_0 + \frac{g}{2} + \frac{F}{3} $

Given, value of uniform acceleration $=a$

Displacement $=s$

Velocity $=v$

By using equation of parabola,

$y=m x^2+c$

and 3rd equation of motion,

$v^2-u^2 =2 as$

If $u =0 \mathrm{~ms}^{-1}$

$\therefore$ $v^2 =2 a s \Rightarrow s=\frac{1}{2 a} v^2+0$

Hence, parabola starts from origin.

Given,

Displacement of particle, $s=9 t^2-2 t^3$

Initial velocity, $u=0 \mathrm{~ms}^{-1}$

Velocity, $v=\frac{d s}{d t}=\frac{d}{d t}\left(9 t^2-2 t^3\right) \Rightarrow v=18 t-6 t^2$

Since, $\quad v=0$

$\Rightarrow 18 t-6 t^2=0 \Rightarrow t=3 \mathrm{~s}$

Given, velocity of car A and B, = 30 km/h

Separation between two cars A and B = 10 km

Let velocity of car C be $v_c$

Time taken, t = 8 min = 8/60 h

Therefore,

$\begin{aligned} &\therefore v_C+30=\frac{A \times B}{t}=\frac{10 \times 60}{8} \\\\ &\Rightarrow v_C+30=75 \\\\ &\Rightarrow v_C=75-30=45 \mathrm{~km} / \mathrm{h}\end{aligned}$

Given, initial speed, u = 36 km/h = 10 ms$^{-1}$

Final speed, v = 0 ms$^{-1}$

Distance travelled, s = 200 m

Let retardation be $a$.

$\begin{aligned} & \because \quad v^2-u^2=2 a s \\ & \therefore \quad a=\frac{v^2-u^2}{2 s}=\frac{0-10^2}{2 \times 200}=-\frac{100}{400}=-0.25 \mathrm{~ms}^{-2} \\ & \therefore \quad \text { Retardation }(a)=0.25 \mathrm{~ms}^{-2} \end{aligned}$

Given, as we know that gravity always pulls the body towards ground and at maximum height, speed of body will be zero, so net acceleration of body will be g towards earth (downward).

As we know that, In case of retardation, the velocity of the body reduces continuously and as the body is moving in a straight line. Therefore, displacement of the body will be equal to distance.

Hence, velocity of body is equal to speed and speed of body will decrease due to retardation.