A car is moving along a straight line,say OP | vedclass

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(B) From the figure,the position of $O$ is $0\; m$,$P$ is $360\; m$,and $Q$ is $240\; m$.Total displacement = Final position - Initial position = $240

Vedclass · Accepted Answer

$10\; m s^{-1}, 20\; m s^{-1}$

Vedclass · Answer

(B) From the figure,the position of $O$ is $0\; m$,$P$ is $360\; m$,and $Q$ is $240\; m$.Total displacement = Final position - Initial position = $240\; m - 0\; m = 240\; m$.Total time taken = $18\; s + 6.0\; s = 24\; s$.Average velocity = $\frac{	ext{Total displacement}}{	ext{Total time}} = \frac{240\; m}{24\; s} = 10\; m s^{-1}$.Total path length = Distance $OP$ + Distance $PQ = 360\; m + (360\; m - 240\; m) = 360\; m + 120\; m = 480\; m$.Average speed = $\frac{	ext{Total path length}}{	ext{Total time}} = \frac{480\; m}{24\; s} = 20\; m s^{-1}$.

$A$ car is moving along a straight line,say $OP$ in the given figure. It moves from $O$ to $P$ in $18\; s$ and returns from $P$ to $Q$ in $6.0\; s$. What are the average velocity and average speed of the car in going from $O$ to $P$ and back to $Q?$

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