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The acceleration of the particles of linear motion changes a = 2s with the displacement(s), and the velocity of the particles is zero when the displacement is zero. Find the corresponding velocity-displacement equation:

 

Option: 1

v=0\ and\ s=0


Option: 2

v=1\ and\ s=\frac{1}{2}


Option: 3

v=2\ and\ s=3


Option: 4

v=4\ and\ s=\frac{2}{3}


Answers (1)

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Acceleration \frac{dv}{dt}=2\times swhere s is displacement and v is velocity at time t;
Let us say that the function of displacement time s is of the following form

s=e^{kt} .....(1)

Now, differentiating equation(1):

\frac{ds}{dt}=ke^{kt}=k\times s\: \: \: \: ..........(2)

Again, differentiating equation (2):

\frac{d^2s}{dt^2}=k^2e^{kt}=k^2s \: \: \: \: .............(3)

The given equation is :

\frac{d^2s}{dt^2}=2\times s \: \: \: \: \: ...........(4)

From equation (3) and(4), we get:

k=\sqrt2 \: \: \: \: \: \: \: .........(5)

From equation (2) we get the velocity–displacement relation:

v=\sqrt2\times s \: \: \: \: \: \: \: \: \: ..........(6)

Therefore, option (1) satisfies the given form which implies v= 0 when s = 0.

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Rishabh

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