Question

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?

Answer 1

@ash2326

Answer 2

@radar @waterineyes

Answer 3

@pinkbubbles

Answer 4

sorry i'm not sure @SolomonZelman

Answer 5

It's ok c:

Answer 6

@Jhannybean !!

Answer 7

@midhun.madhu1987 @mathstudent55

Answer 8

@ParthKohli

Answer 9

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: $$d=\left\|r\left(\frac1{2a}\right)-r(0)\right\|=\|r_0\|\cdot\frac1{2a}\left(1-\frac12\right)=\frac{\|r_0\|}{4a}$$