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#### The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s−1.  At, t=0 the displacement is 5 m.  What is the maximum acceleration ?  The initial phase IS Option: 1 500 m/s2   Option: 2 $500\sqrt{2} \ m/s2$ Option: 3 750 m/s2   Option: 4$750\sqrt{2} \ m/s2$ m/s2

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#### A particle is moving along the x- axis with its coordinate with time 't' given by $x(t)=10+8t-3t^2$. Another particle is moving along the y-axis with its coordinate as a function of time given by $y(t)=5-8t^3.$ At $t=1$ s, the speed of the second particle measured in the frame of the first particle is given by $\sqrt{v}$. Then $v$ (in m/s) is ____________. Option: 1 580 Option: 2 700 Option: 3 100 Option: 4 300

\begin{aligned} &x_{A}=-3 t^{2}+8 t+10 \\ &\vec{v}_{A}=(-6 t+8) \hat{i} \\ &\text { At time } t=1 \mathrm{~s} \\ &\vec{v}_{A}=2 \hat{i} \\ &Y_{B}=5-8 t^{3} \end{aligned}

$\vec{v}_{B}=-24 t^{2} j$

At time t= 1 s

$\vec{v_{B}}=-24 \hat{j}$

\begin{aligned} &\sqrt{v}=\left|\vec{v}_{B}-\vec{v}_{A}\right|=|-24 \hat{j}-2 \hat{i}| \\ &\sqrt{v}=\sqrt{580} \end{aligned}

So the value of v is 580

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#### A particle starts from the origin at t=0 with an initial velocity of $3.0\hat{i}\; m/s$ and moves in the x-y plane with a constant acceleration $\left ( 6.0\hat{i}+4.0\hat{j} \right )m/s^{2}.$ The x-coordinate of the particle at the instant when its y-coordinate is 32 m is D meters. The value of D is : Option: 1 60 Option: 2 50 Option: 3 32 Option: 4 40

$\\ \text{Given}:\\ \text{Taking motion along y -axis} \\ \text{Displacement along y-axis}(s_{y})=32 \ m\\ \text{Acceleration along y-axis},(a_{y})=4\ m/s^{2}\\ \text{Initial velocity along y-axis},(u_{y})=0 \ m/s\\\\ \text{From 2nd equation of motion}, \\ s_{y}=u_{y}t+\frac{1}{2}a_{y}t^{2} \\ \\ \Rightarrow 32=0(t)+\frac{1}{2}(4)(t^{2})\\ \\ \Rightarrow 32=2t^{2}\\ \\\Rightarrow t=4\ s$

\\ \text{Now taking motion along x- axis.} \\ \text{Displacement along x-axis} \left(\mathrm{s}_{\mathrm{x}}\right)= ? \\ \text{Acceleration along x-axis },(a_{x})=6 \ m/s^{2} \\ \text{initial velocity along x-axis} \left(\mathrm{u}_{\mathrm{x}}\right)=3 \ m/s \\ \begin{aligned} &\text { So, } \mathrm{S}_{\mathrm{x}}=\mathrm{u}_{\mathrm{x}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{x}} \mathrm{t}^{2} \\ &=3 \times 4+\frac{1}{2} \times 16 \\ &=12+48 \\ &=60 \mathrm{~m} \end{aligned}

#### A ball is dropped from the top of a 100 m high tower on a planet. In the last $\frac{1}{2}$ s before hitting the ground, it covers a distance of 19 m. Acceleration due to gravity ( in ms-2) near the surface on that planet is _______. Option: 1 8 Option: 2 4 Option: 3 2 Option: 4 6

Let the total time taken by the particle is t

So distance covered by a particle in t sec=$S_t=ut+\frac{1}{2}gt^2=\frac{1}{2}gt^2$

Similarly, distance covered by particle in $(t-\frac{1}{2})$ sec=$S_{ t-\frac{1}{2}}=\frac{1}{2}g( t-\frac{1}{2})^2$

So  distance covered by particle in last (1/2) sec of its journey=$S_t-S_{ t-\frac{1}{2}}$

So $19=S_t-S_{ t-\frac{1}{2}}=\frac{1}{2}gt^2-\frac{1}{2}g( t-\frac{1}{2})^2 =\frac{1}{2}g(t-\frac{1}{4})....(1)$

and $S_t= \frac{1}{2}gt^2=100..(2)$

from equations (1) and (2)

we get $19t^2-100t+25=0\Rightarrow t=5 , or \ \ t=0.26$

putting t=5 we get g = 8 m/s2

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#### The distance $x$ covered by a particle in one dimensional motion varies with time t as . If the acceleration of the particle depends of $x$ as $x^{-n}$, where n is an integer, the value of n is ______ Option: 1 3 Option: 2 4 Option: 3 1 Option: 4  0

$x^2 = at^2+2bt + c \ \ \ - \left ( 1 \right )$

$\text{Differentiating equation 1 with respect to 't', we get }$

$2x\frac{\text d x}{\text dt} = 2at + 2b \ \ \ \ -\left ( 2 \right )$
$\text{Again differentiating equation 2 with respect to 't', we get }$

$\begin{gathered} 2\left[x \frac{d^{2} x}{d t^{2}}+\frac{d x}{d t} \cdot \frac{d x}{d t}\right]=2 a \\ x \frac{d^{2} x}{d t^{2}}+\left(\frac{dx}{d t}\right)^{2}=\frac{2 a}{a} \end{gathered} \\ \text{Put the value of}\ \frac{d x}{d t}\ \text{from eq.} (2) \text{ we have} \\ x \frac{d^{2} x}{d t^{2}}+\left(\frac{2 a t+2 b}{2 x}\right)^{2}=2\\ \\ \frac{d^{2} x}{d t^{2}}=-\frac{(2 at+2 b)^{2}}{4 x^{2} x}=\frac{-(2 at+2 b)^{2}}{4 x^{3}}\\ \\ or \frac{d^{2} x}{d x^{2}} \propto \frac{A}{x^{3}}\left\{\begin{array}{l}A \text { is constant here i.e }A=\frac{(-2at+2b)^{2}}{4} \\ \\ \end{array}\right\}\\ \\\frac{d^{2}x}{dt^{2}} \propto x^{-3}$

$\therefore n=3$

#### A particle moves such that its position vector $\widehat{r}\left ( t \right ) = cos\omega t\widehat{i} + sin \omega t\widehat{j}$ where $\omega$ is a constant and t is time. Then which of the following statements is true for the velocity $\vec{v}(t)$ and acceleration $\vec{a}(t)$ of the particle :  Option: 1 and both are parallel to Option: 2 is perpendicular to and is directed away from the origin Option: 3  and both are perpendicular to Option: 4  is perpendicular to and is directed towards the origin

$\\\vec r=cos\omega t\ i+ sin\omega t\ j\\\vec v=\frac{d\vec r}{dt}=-\omega sin\omega t i+\omega cos\omega t j$

$\\\vec a=\frac{d\vec v}{dt}=-\omega ^2(cos\omega t\ i+ sin\omega t\ j)=-\omega^2 \vec r$

implies a is anti parallel to r

$\\\vec v. \vec r=0\\$

Implies v is perpendicular to r

hence the correct option is (4).

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#### The sum of two forces  $\vec{P}$ and $\vec{Q}$ is  such that $\left | \vec{R} \right |=\left | \vec{P} \right |.$ The angle $\theta$ (in degrees) that the resultant of $2\vec{P}$ and $\vec{Q}$ will make with  $\vec{Q}$ is , Option: 1 90 Option: 2 120 Option: 3 45 Option: 4 30

As

$\\ |\vec{P}+\vec{Q}|=|\vec{P}| \\ \because |\vec{R}|=|\vec{P}| \\\Rightarrow P^2+Q^2+2PQcos\theta =P^2\\ \Rightarrow Q+2Pcos\theta =0...(1)$

Now Vector P is doubled but vector Q is the same so the angle between $2\vec{P} \ and \ \vec{Q}$ will be also $\theta$

And If resultant of $2\vec{P} \ and \ \vec{Q}$ is $\vec{R'}$

Then from the figure angle between $\vec{R'} \ and \ \vec{Q}$ is $\alpha$

and $tan \alpha =\frac{2Psin\theta }{Q+2Pcos\theta }\\ From \ equation \ (1) \\ \text{we get tan} \alpha =\infty\Rightarrow \alpha =90^0$

#### A car accelerates from rest at a constant rate for sometime after which it decelerates at a constant rate to come to rest. If the total time elapsed is t seconds, the total distance travelled is :Option: 1Option: 2Option: 3Option: 4

Let the car accelerate for time $t_{1}$ traveling distance $S_{1}$ and acquire maximum velocity v.
Then,$S_{1}=\frac{1}{2} \alpha t_{1}^{2}$ [from the equation, $S=u t+\frac{1}{2} a t^{2}$ ]
and $v=\alpha t_{1} \quad$  [from the equation, v=u+a t ]

$\Rightarrow \quad S_{1}=\frac{1}{2} \alpha \times \frac{v^{2}}{\alpha^{2}} =\frac{v^{2}}{2 \alpha} ......(1)$
After this car decelerates for time $t_{2}$ to come to rest
Hence, $\quad S_{2}=v t_{2}-\frac{1}{2} \beta t_{2}^{2}$       [from the equation, $S=u t-\frac{1}{2} a t^{2}$ ]
and $0=v-\beta t_{2}$         [from the equation, u=v-a t ]
$\Rightarrow \quad S_{2}=\frac{v^{2}}{\beta}-\frac{v^{2}}{2 \beta}=\frac{v^{2}}{2 \beta} \ldots (ii)$

$\begin{array}{lrl}\text { Now, } & t_{1}+t_{2} & =t \\ \\ \Rightarrow & \frac{v}{\alpha}+\frac{v}{\beta}=t \\ \\ \Rightarrow & v=\left(\frac{\alpha \beta}{\alpha+\beta}\right) t\end{array}$

Also, total distance traveled,

\begin{aligned} S &=S_{1}+S_{2} \\ &=\frac{v^{2}}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) \\ &=\frac{1}{2}\left(\frac{\alpha \beta t}{\alpha+\beta}\right)^{2}\left(\frac{\alpha+\beta}{\alpha \beta}\right) \\ &=\frac{1}{2}\left(\frac{\alpha \beta}{\alpha+\beta}\right) t^{2} \end{aligned}

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#### A swimmer wants to cross a river from point to point . Line makes an angle of with the flow of river. Magnitude of velocity of the swimmer is same as that of the river. The angle with the line should be _______swimmer reaches point

${\bar{V}_{M / G}}={\bar{V}_{M / R}}+{\bar{V}_{R}}$

$\bar{V}_{M / G} \rightarrow Velocity \: of \: swimmer \: with \: ground$

$V_{M/G} must \: be \: along \: A B to reach pt. B$

\begin{aligned} V_{M / G}=V_{M / R} \cos \theta+V_{R} \cos 30^{\circ} \\ \end{aligned}

\begin{aligned} V_{M / R} \sin \theta &=V_{R} \sin \left(30^{\circ}\right) \\ V_{M / R} &=V_{R}(\operatorname{Given}) \\ \sin \theta &=\sin 30^{\circ} \\ \theta &=30^{\circ} \end{aligned}

#### The magnitude of vectors  in the given figure are equal. the direction of  with x-axis will be:Option: 1Option: 2Option: 3Option: 4

$let\, \left |\bar{OA} \right |= \left |\bar{OB} \right |= \left |\bar{OC} \right |= a$
$\bar{OA}= a\cos 30^{\circ}\hat{i}+a\sin 30^{\circ}\hat{j}$
$\bar{OB}= a\cos 60^{\circ}\hat{i}+a\sin 60^{\circ}\left ( -\hat{j} \right )$
$\bar{OC}= a\cos 45\left ( \hat{i} \right )+a\sin 45\left ( \hat{j} \right )$
$\bar{OA}+\bar{OB}-\bar{OC}=\left ( \frac{\sqrt{3}a}{2}+\frac{a}{2}+\frac{a}{\sqrt{2}} \right )\hat{i}+\left ( \frac{a}{2}-\frac{\sqrt{3}a}{2}-\frac{a}{\sqrt{2}} \right )\hat{j}$

$\tan \theta= \frac{R_{y}}{R_{x}}$
$= \frac{\left ( \frac{a}{2} -\frac{\sqrt{3}a}{2}-\frac{a}{\sqrt{2}}\right )}{\left ( \frac{\sqrt{3}a}{2}+\frac{a}{2}+\frac{a}{\sqrt{2}}\right )}$
$\tan \theta= \frac{\left ( 1-\sqrt{3}-\sqrt{2} \right )}{\left (\sqrt{3}+1+\sqrt{2} \right )}$
The correct option is (2)