Physics, asked by rajeshchaudhari1616, 1 month ago

Joseph jogs from 1 end to other end b of a straight road 300 m in 2 min 30 sec and then turns around and jogs 100 m back to point c in another 1 minute what are the average speed and velocity in jogging​

Answers

Answered by Jesscameron
1

Explanation:

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

Answered by iTzpyaribachi
2

Velocity = dispacement / time

Speed = distance / time

a) when he jogs from A to B on a straight road,

displacement = distance = 300m

time = 2 minutes 30 seconds = 150 s

velocity = 300/150 = 2 m/s

speed = 300/150 = 2m/s

b)when he jogs from A to B and turns back to C,

displacement = 300-100 = 200m

distance = 300+100 = 400m

time = 3 minute 30 second = 210 s

velocity = 200/210 = 20/21 m/s

speed = 400/210 = 40/21 m/s

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