## Abhisar one year ago position vector of a particle is given by $$\sf r = r_0(1-at)t$$ , where t is the time and a as well as $$\sf r_0$$ are constant. How much distance is covered by the particle in returning to the starting point?

1. Abhisar

@ash2326

2. Abhisar

3. anonymous

@pinkbubbles

4. anonymous

sorry i'm not sure @SolomonZelman

5. Abhisar

It's ok c:

6. Abhisar

@Jhannybean !!

7. Abhisar

8. Abhisar

@ParthKohli

9. anonymous

so you want to solve for when $$r(t)=0\implies t-at^2=1\implies t(at-1)=0$$ so $$t=0\text{ or }t=\frac1a$$ so presumably we're taking the $$t=1/a$$ root, and the distance covered is given by: \begin{align*}s&=\frac12\int_0^t \|r'(\tau)\|\,d\tau\end{align*} i.e. half the total distance, but notice our path is completely collinear, and we know we reach our furthest point at $$t=1/(2a)$$ (midpoint of the roots) so our distance on the return is equal to the distance going: $$d=\left\|r\left(\frac1{2a}\right)-r(0)\right\|=\|r_0\|\cdot\frac1{2a}\left(1-\frac12\right)=\frac{\|r_0\|}{4a}$$