Question

As a train accelerates away from a station, it reaches a speed of 4.6 m/s in 5.2 s. If the train's acceleration remains constant, what is its speed after an additional 7.0 s has elapsed? Express your answer using two significant figures.

Answer 1

Answer:

Vf = 10.76 m/s

Explanation:

Train kinematics

The train moves with uniformly accelerated movement

$V_f = V_o + a*t$ Formula (1)

Vf: Final speed (m/s)

V₀: Inital speed (m/s)

t: time in seconds (s)

a: acceleration (m/s²)

Movement from t = 0 to t = 5.2s

We replace in formula (1)

4.6 = 0 + a*5.2

a = 4.6/5.2 = 0.88 m/s²

Movement from t = 5.2s to t = 5.2s + 7s = 12.2s

We replace in formula (1)

$V_f = 4.6 + 0.88*7$

Vf = 10.76 m/s