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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 \frac{\pi }{4}
Option: 1 500 m/s2  
Option: 2 500\sqrt{2} \ m/s2
Option: 3 750 m/s2  
Option: 4750\sqrt{2} \ m/s2 m/s2  

As learnt in 


f_{max}=\overset{2}{\omega}A, \:\:\:V_{max}=a \omega


X=a\: \sin \left (\omega t+ \frac {\pi}{4}\right ) [given]\:\:\:\:\:\:\:\:\:\:[omega =10]

At f=0

5=a\:\sin \frac{\pi}{4}                                        a= 5 \sqrt2

Maximum acceleration = \omega^{2} \times a

=100 \times 5 \sqrt 2

=500 \sqrt 2


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Posted by

vishal kumar

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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vishal kumar

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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}


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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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vishal kumar

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The distance x covered by a particle in one dimensional motion varies with time t as x^2 = at^2+2bt + c. 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

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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 \vec{v} and \vec{a} both are parallel to \vec{r}
Option: 2 \vec{v}is perpendicular to \vec{r}and \vec{a} is directed away from the origin
Option: 3 \vec{v} and \vec{a} both are perpendicular to \vec{r}
Option: 4 \vec{v} is perpendicular to \vec{r} and \vec{a} 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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Posted by

vishal kumar

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The sum of two forces  \vec{P} and \vec{Q} is \vec{R} 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


 \\ |\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

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Ritika Jonwal

A car accelerates from rest at a constant rate \alpha for sometime after which it decelerates at a constant rate \beta to come to rest. If the total time elapsed is t seconds, the total distance travelled is :
Option: 1 \frac{4\; \alpha \; \beta }{\left ( \alpha +\beta \right ) }\; t^{2}
Option: 2 \frac{2\; \alpha \; \beta }{\left ( \alpha +\beta \right ) }\; t^{2}
Option: 3 \frac{ \alpha \; \beta }{2\left ( \alpha +\beta \right ) }\; t^{2}  
Option: 4 \frac{ \alpha \; \beta }{4\left ( \alpha +\beta \right ) }\; t^{2}

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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Kuldeep Maurya

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A swimmer wants to cross a river from point A to point \mathrm{B}. Line \mathrm{AB} makes an angle of 30^{\circ} with the flow of river. Magnitude of velocity of the swimmer is same as that of the river. The angle \theta with the line \mathrm{AB} should be _______^{0},swimmer reaches point B.

{\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}

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vishal kumar

The magnitude of vectors \vec{OA},\vec{OB}\, and\, \vec{OC} in the given figure are equal. the direction of  \vec{OA}+\vec{OB}-\vec{OC} with x-axis will be:

Option: 1 \tan^{-1}\frac{\left ( \sqrt{3}-1+\sqrt{2} \right )}{\left ( 1+\sqrt{3}-\sqrt{2} \right )}
Option: 2 \tan^{-1}\frac{\left (1- \sqrt{3}-\sqrt{2} \right )}{\left ( 1+\sqrt{3}+\sqrt{2} \right )}
Option: 3 \tan^{-1}\frac{\left ( \sqrt{3}-1+\sqrt{2} \right )}{\left ( 1-\sqrt{3}+\sqrt{2} \right )}
Option: 4 \tan^{-1}\frac{\left (1+ \sqrt{3}-\sqrt{2} \right )}{\left ( 1-\sqrt{3}-\sqrt{2} \right )}

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)

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vishal kumar

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