A particle moving in a straight line has velocity-displacement equation as $v=5\sqrt {1+s}$. Select the correct alternative.

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a) Particle is initially at rest.

b) Initial velocity of the particle is 5m/s and particle has a constant acceleration of 12.5m/s^2

c) Particle moves with a uniform velocity

d) None of these

Correct answer is b

Since $v=\frac{ds}{dt}$ $$dt=\frac{ds}{v}$$ $$t=\frac{2\sqrt{1+s}}{5} + C$$ at initial velocity, t=0.

And I don’t know what to do next. Please help.

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Let the initial displacement $s(0)=0$ in $v=5\sqrt {1+s}$ to get the initial velocity $v(0)=5m/s$.

For acceleration, take the time derivatives on both sides of

$$v=5\sqrt {1+s}\tag{1}$$

to get

$$\frac{dv}{dt}=\frac 52\frac{1}{\sqrt{1+s}}\frac{ds}{dt}\tag{2}$$

Plugging (1) into (2) and recognizing $v=ds/dt$, (2) becomes,

$$\frac{dv}{dt}=\frac {5}{2}\frac{1}{\frac 15 v}v=\frac{25}{2}=12.5 \tag{2}$$

which shows that the particle has a constant acceleration of $12.5m/s^2$. Thus, the correct answer is (b).