Question

A jogger jogs from one end to other straight 300 m road same from point A to point B in 2.5 minutes and then turn around and jogs 100 m ack towards the starting point C in another 1 minute.What is the joggers average speed and velocity in jogging from point A to point B.

Answer 1

Speed=Total distance/Total time taken
Velocity=displacement/Total time taken

NOTE: "displacement" is the minimum distance from the initial and the final point,whereas "distance" is the total distance.

#FIRST PART:FINDING THE SPEED FROM MOVING A TO B

Total distance covered in moving from A to B = 300m,
Total time taken to cover this distance= 2.5min = 150sec (since the unit of speed should be m/s)
Speed=300/150 = 2m/s

#SECOUND PART:FINDING THE VELOCITY FROM MOVING A TO B:

Velocity=displacement/time taken
            =300m/150s=2m/s   [displacement=distance,since minimum distance from A to B is also 300]



Since you have given the distance and time taken for reaching point C,i guess your question is incomplete, 

#THIRD PART:TO FIND THE VELOCITY IS MOVING FROM A TO C (Assuming this is the question)

Velocity=displacement/time taken
 [displacement=300-100=200(since the jogger jogs back 100 m, his distance between initial and final position decreases by 100)]

Total time taken for the journey=2.5+1=3.5min=210s

Velocity=200/210 =0.9523m/s ≈ 1m/s (approx)