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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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Aditya
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  • You can't make it any further. You need to know the initial position. – Mostafa Ayaz Sep 06 '19 at 13:38
  • Is your answer supposed to be an equation of $s_0$? You could solve for $C$ in terms of $s_0$ by making $t=0$, then solve for $s$ in terms of $t$ and $s_0$. Then take the derivative of that equation with respect to $t$. – Joe Sep 06 '19 at 13:42
  • I get $v_0=5\sqrt{1+s_0}$, which I see now is obvious. – Joe Sep 06 '19 at 13:49
  • Updated the question. I thought that that part would be enough. Anyway, the original question is up there. – Aditya Sep 06 '19 at 13:50
  • If you do what I suggested in my comment, you can solve for $s$ in terms of $t$ and $s_0$, then take two derivatives to see that the acceleration is constant and has that value. However, I don’t think you can know the initial velocity is 5 without knowing that the particle starts at $s=0$. – Joe Sep 06 '19 at 13:55
  • Yeah, I don’t think I will get an answer with that. – Aditya Sep 06 '19 at 14:18
  • By “don’t think [you] will get an answer”, are you saying that you agree that you need $s_0$ to know $v_0$, or are you saying that you don’t know how to do what I described in my comment? – Joe Sep 06 '19 at 14:50
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    Since $s$ is called “displacement,” it looks like you’re meant to assume that $s_0=0$. – amd Sep 06 '19 at 19:28

1 Answers1

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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).

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