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Abhisar
 one year ago
position vector of a particle is given by \(\sf r = r_0(1at)t\) , 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?
Abhisar
 one year ago
position vector of a particle is given by \(\sf r = r_0(1at)t\) , 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?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry i'm not sure @SolomonZelman

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0@midhun.madhu1987 @mathstudent55

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you want to solve for when \(r(t)=0\implies tat^2=1\implies t(at1)=0\) 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: $$d=\left\r\left(\frac1{2a}\right)r(0)\right\=\r_0\\cdot\frac1{2a}\left(1\frac12\right)=\frac{\r_0\}{4a}$$
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