Question

Joseph jogs from one end A to the other end B of a straight 300m road in 2min.30 s and then turnsaround and jogs 100m back to point C in another 1min. What are Joseph’s average speeds and velocitiesin jogging from A to B and from A to C?​

Answer 1

Explanation:

ANSWER

From pont A to B

Distance covered=300m

Displacement=300m

Time taken=2 min 30 sec=(2×60)+30=150 sec

Average speed=

Timetaken

Distance covered

=

150

300

=2 m/sec

Average velocity=

Timetaken

Displacement

=

150

300

=2 m/sec

From point A to C

Distance covered=300+100=400m

Displacement=300−100=200m

Time taken=3 min 30 sec=(3×60)+30=210 sec

Average speed=

Timetaken

Distance covered

=

210

400

=1.90 m/sec

Average velocity=

Timetaken

Displacement

=

210

200

=0.95 m/sec