Question

A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from the top; 2.00 s after that it is 25.6 m from the top, 2.00 s later 40.0 m from the top, and 2.00 s later it is 57.6 m from the top.What is the magnitude of the average velocity of the sled during each of the 2.00-s intervals after passing the 14.4-m point?

Answer 1

Answer:

V₁  = 5.6 m/s

V₂ = 7.2 m/s

V₃ = 8.8 m/s

Explanation:

Average velocity: Average velocity can be defined as the ratio of the total  displacement to the total time taken. The S.I unit of Average velocity is m/s.

For the first 2 s,

V₁ = Δd₁/t

Where V₁  = Average velocity for the first 2 s

Where Δd₁= distance, t = time

Δd₁ = 25.6-14.4 = 11.2 m t = 2 s

V₁ = 11.2/2

V₁ = 5.6 m/s

For the second 2 s,

V₂ =Δd₂/t

Where V₂ = average velocity for the second 2 s.

Δd₂= 40-25.6 = 14.4 m, t= 2 s

V₂ = 14.4/2

V₂ = 7.2 m/s

For the last 2 seconds,

V₃ =Δd₃/t

Where V₃ = average velocity for the last 2 s

where Δd₃ = 57.6- 40 = 17.6 m, t = 2 s

V₃ = 17.6/2

V₃ = 8.8 m/s.