Question

2. Joseph jogs from one end A to the other end B of a straigh
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 As
B and (b) from A to C​

Answer 1

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

Answer 2

QUESTION -

Joseph jogs from one end A to the other end B of a straight 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

ANSWER -

(1)

average speed from A to B

In this case the total distance travelled by the object will be AB = 300m

total time taken by the body to go form A to C = time from A to B = 2min50s = 170s

so,

average speed = 300/170 = 1.764 m/s

similarly, average velocity from A to B

here the net displacement acquired will be = AB = 300m

total time taken will be the same as previously = 170s

thus,

average velocity = 300/170 = 1.764 m/s

(2)

average speed from A to C

In this case the total distance travelled by the object will be AC = AB + BC = 300m + 100m = 400m

total time taken by the body to go form A to C = time from A to B + time from B to C = 2min50s + 1min = 3min50s = 230s

so,

average speed = 400/230 = 1.739 m/s

similarly, average velocity from A to C

here the net displacement acquired will be AC = AB - BC = 300m - 100m = 200m

total time taken will be the same as previously = 230s

thus,

average velocity = 200/230 = 0.869 m/s