A body moves from point A to point B with a velocity of 5m/s and returns to point A with a speed of 3m/s what is the average velocity?

(Technically, the average velocity is zero because the start & end points are the same, and the start & end speeds are the same (zero). The average *speed* is what is wanted here.)

Let the distance from A to B be x meters.
The time needed to go from A to B is (x m)/(5 m/s) = x/5 s.
The time needed to go from B to A is (x m)/(3 m/s) = x/3 s.
The total time is then x/5 + x/3 = 8x/15 s.
The total distance is 2x m, so the average speed is (2x m) / (8x/15 s) = 15/4 m/s = 3.75 m/s.

Average velocity = 0 m/s
Average speed = 3.75 m/s


====
(A) To find the average velocity :

The body moves from A to B, then from B to A.
Displacement = 0 m

Average velocity = Displacement / Time = (0 m) / Time = 0 m/s


====
(B) To find the average speed :

Let d m be the distance between A and B.

Total distance travelled = 2d m

Time taken from A to B = (d m) / (5 m/s) = d/5 s
Time taken from B to A = (d m) / (3 m/s) = d/3 s
Total time taken = [(d/5) + (d/3)] s = [(3d/15) + (5d/15)] s = 8d/15 s

Average speed = Distance / Time = (2d m) / (8d/15 s) = 3.75 m/s

(5+3)/2=4m/s

30/8=3 6/8 m/s

 
♦ The vectorial average speed is 0↑.
♦ The scalar average :
<V> = 2.D/T = 2.D/(T1 + T2) = 2.D/(D/V1 + D/V2)
<V> = 2/(1/V1 + 1/V2) = 2/(1/5 + 1/3) = 3.75 m/s

60.