Physics, asked by ranokang8084, 8 months ago


2. Joseph jogs from one end A to the other end B of a
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?

Answers

Answered by ashimaarorapurvika
13

Answer:

From pont A to B

Distance covered=300m

Displacement=300m

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

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

Explanation:

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