Physics, asked by mngaur2, 1 year ago

Joseph jogs from one end A to the other end B of a straight 300m road in 2 minutes 30 seconds and then turns around and jogs 100m back to point C in another 1 minute. What are Joseph's average speeds and velocities in jogging (a) from A to B (b) from A to C?

Answers

Answered by shub79
16

Distance from A to B = 300m

Displacement from A to B = 300m

Time taken to go from A to B = 2 min 50sec =170sec

a) Average speed in jogging from A to B = distance ÷ time

300m÷170sec = 1.76ms raised to power -1

Average velocity in jogging from A to B = displacement ÷ time

300m ÷170sec = 1.76ms

b) Distance covered from A to C = 300m +100m = 400m

Displacement from A to C = 300-100= 200

Time taken to go from A to C = 170 sec + 60s = 230s

Average speed from A to C = distance ÷time

400÷230= 1.74ms raised to power -1

Average velocity = displacement ÷ time

200÷230= 0.87 ms raised to power -1.


mngaur2: Time taken to go from a to b is 2 minutes ep seconds
Answered by Anonymous
10

Answer:

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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