Question

An object thrown straight up a distance ymaxymax. After tt seconds, it falls back and is caught again, just as it reaches the height from which it was thrown. Claudia says its average velocity was zero, and Hossein says its average speed was 2ymaxt2ymaxt . What would you say to help them out?

Answer 1

Answer:

Avg.velocity=(Δy/ Δt)  =(net displacement/ total time for journey)

Δy = 0

Δt = t

so avg. velocity = 0/t =0

Avg. speed =(total distance traveled/ total time for journey)

total distance = up +down = Ymax+Ymax= 2 Ymax

total time = t

avg. speed = 2 Ymax/t

Explanation:

Since there is no net displacement from the original position,velocity is zero. Claudia is right!

while it covered some distance in time t so its speed is not as qouted by Hossien