A train started from rest and moved with constant acceleration. At one time it was traveling 27 m/s, and 150 m farther on it was traveling 54 m/s. Calculate (a) the acceleration, (b) the time required to travel the 150 m mentioned, (c) the time required to attain the speed of 27 m/s, and (d) the distance moved from rest to the time the train had a speed of 27 m/s.

Explanation:

(a) Given:

Δx = 150 m

v₀ = 27 m/s

v = 54 m/s

Find: a

v² = v₀² + 2aΔx

(54 m/s)² = (27 m/s)² + 2a (150 m)

a = 7.29 m/s²

(b) Given:

Δx = 150 m

v₀ = 0 m/s

a = 7.29 m/s²

Find: t

Δx = v₀ t + ½ at²

150 m = (0 m/s) t + ½ (7.29 m/s²) t²

t = 6.42 s

(c) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: t

v = at + v₀

27 m/s = (7.29 m/s²) t + 0 m/s

t = 3.70 s

(d) Given:

v₀ = 0 m/s

v = 27 m/s

a = 7.29 m/s²

Find: Δx

v² = v₀² + 2aΔx

(27 m/s)² = (0 m/s)² + 2 (7.29 m/s²) Δx

Δx = 50 m

because if the tool that you are using is not balanced it there would be false mass.