Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Given:
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute.
To do:
To find Average speeds and velocities in jogging-
$(a)$ from A to B
$(b)$ from A to C
Solution:
$(a)$. From A to B:
Total distance from A to B $=300\ m$
Because the road is straight and Joseph jogs in a single direction, so
in this case, net displacement will be $300\ m$.
Total time taken to cover the distance from A to B, $=2\ minutes\ 30\ secomds=2\times {60}+30\ s=150\ s$
Therefore, the Average speed in jogging from $A\ to\ B$ is given as-
Average Speed$=\frac {Total\ distance\ travelled}{Total\ time\ taken}$
$=\frac {300}{150}$
$=2\ m/s$
Average Velocity$=\frac {Net\ displacement}{Total\ time\ taken}$
$=\frac {300}{150}$
$=2\ m/s$
$(b)$. From A to C:
Total Distance covered from A to C $=$ Distance between A to B $+$ Distance between B to C
$=300\ m+100\ m=400\ m$
Net displacement $=300\ m-100\ m=200\ m$
Total time taken from A to C $=$ Time from A to B + Time from B to C
$=150\ s+60\ s=210\ s$
Therefore, average speed $=\frac {Total\ distance\ travelled}{Total\ time\ taken}$
$=\frac {400}{210}$
$=1.9\ m/s$
Therefore, the Average velocity $=\frac {Net\ displacement}{Total\ time\ taken}$
$=\frac {200}{210}$
$=0.95\ m/s$
Thus, the Average speed while traveling from A to B = 2 m/s.
Average velocity while traveling from A to B = 2 m/s.
Average speed while traveling from A to C = 1.9 m/s.
Average velocity while traveling from A to C = 0.95 m/s.
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