Question

A body with a uniform acceleration travels distances of 24m and 64m during the first two equal consecutive intervals of time, each of duration 4s. Determine the initial velocity and acceleration of the moving body.​

Answer 1

Answer:

Explanation:

Average velocity in the 24 m interval is 24 / 4 = 6 m/s

Average velocity in the 64 m interval is 64 / 4 = 16 m/s

There is a 4 second interval between the two points where average velocity equals actual velocity

a = Δv/t = (vf - vi) / t = (16 - 6) / 4 = 2.5 m/s²

s = v₀t + ½at²

24 = v₀(4) + ½(2.5)4²

4v₀ = 24 - 20

v₀ = 1 m/s

Not asked for but the velocity at the end of the first segment and beginning of the second segment is 11 m/s and final velocity is 21 m/s