Question

A person moves 100m from A to B in 5 seconds and travel the same distance returning in 10 second what is average velocity??

Answer 1

Let the distance from AA to BB be xx ,time taken be t1t1 and t2t2 and speeds be v1v1 and v2.v2.

So, t1=x/v1t1=x/v1————-(i)

And, t2=x/v2t2=x/v2 ———(ii)

Now, according to the definition of averageaverage speed=speed= TotalTotal distance/distance/ totaltotal timetime

So, Vavg=Vavg= 2x/[(x/v1)+(x/v2)]2x/[(x/v1)+(x/v2)]

Simplifying this you can write :-

Vavg=Vavg= 2x/[x(v1+v2)/v1v2]2x/[x(v1+v2)/v1v2]

So, Vavg=2v1v2/(v1+v2)Vavg=2v1v2/(v1+v2)

Putting the values for v1=20m/sv1=20m/sandv2=30m/sv2=30m/s in the equation we can write :-

Vavg=(2×30×20)/50=24m/s.Vavg=(2×30×20)/50=24m/s.

In case of average velocity it is 0 m/s because the displacement is 0 as the body returns to its original position.

Answer 2

displacement = 0

total time = 5 + 10 = 15

average velocity = 0/15 = 0