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?

AcademicPhysicsNCERTClass 9

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.

raja
Updated on 10-Oct-2022 13:22:18

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