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Then it’s location after 7 seconds will be the same as that after-?

Now my super-efficient and productive brain (#sarcasm) can figure out that it’s 3 seconds, since it has got a constant velocity and 10-7=3. But I doubt future questions like this will be that easy. Is there anyway to solve it ‘properly’?

Aditya
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1 Answers1

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No, you are given that the acceleration is constant. The velocity is then a linear function of time and the position is quadratic. The equation for the velocity and position when the acceleration is constant is $$v=v_0+at\\s=s_0+v_0t+\frac 12at^2$$ You should be able to derive these by integration. $s_0$ is the position at $t=0$ and $v_0$ is the velocity at $t=0$. We are given $s(10)=s(0)=s_0$. By taking the derivative you can show that $v(5)=0$, so $v_0=-5a$ and plugging in you can find that your intuition is correct.

Ross Millikan
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  • Yeah I am sorry. I basically thought of it as $\frac{dv}{dt}=K$ so assumed (wrongly) that velocity was constant. – Aditya Sep 03 '19 at 17:09