A boy jogs from one end A to the other end B of a straight 300m road in 2min 30sec and then turns back and jogs 100m back to point C in another 1min. Calculate boy's average speed and average velocity (a) during first 2min 30sec (b) during the whole journey
Answers
A.AVERAGE SPEED=TOTAL DISTANCE COVERED / TOTAL TIME TAKEN.
DURING FIRST 2MIN30SEC
S=300M
T=2 MIN 30 SEC=150 SEC
V=S/T
V=300/150
V=2M/S
B.DURING THE WHOLE JOURNEY
AVERAGE SPEED=TOTAL DISTANCES COVERED/TOTAL TIME INTERVALS
S1=300M
S2=100M
T1=150 SEC
T2=60 SEC
V=S1+S2/T1+T2
V=400/210
V=1.9M/S
A.AVERAGE VELOCITY=TOTAL DISPLACEMENTS COVERED / TOTAL TIME TAKEN
V=S/T
V=300/150
V=2M/S
B.AVERAGE VELOCITY=TOTAL DISPLACEMENT/TOTAL TIME
TOTAL DISPACEMENT=300-100 [AS HE IS RETURNING 100M BACK]=200
TOTAL TIME = 150+60 = 210SEC
V=S/T
V=200/210
V=0.95M/S
a) During first 2min 30sec [From A to B]
Average Speed from A to B = (Total distance)/(Total time)
Given, Distance = 300 m and time = 2 min 30 sec = 2*60 +30 = 150 sec
Average speed = 300/150 = 2 m/s
Now,
Average velocity from A to B (AB) = Dispalcemet/Time
Average velocity = 300/150 = 2 m/s
b) During the whole journey [From A to C]
Now, total distance covered from A to C (AC) = AB + BC = 300 + 100 = 400 m
Given time = 1 min
Time taken from A to C = 150 sec + 1 min = 150 sec + 60 sec = 210sec
Average speed from A to C = (Total distance)/(Total time)
Average speed = 400/210 = 1.904 m/s
Average velocity from A to C = Displacement/time
Average velocity = (AB - BC)/time = (300 - 100)/210
= 200/210 = 0.952 m/s