Question

A train from at rest gains speed of 72 km/h in 5 minutes .calculate its accelaration and distance

Answer 1

Answer:

72 km / h = 72000 m / 60 × 60 s

⇒20 m / s

5 min = 5 × 60 s

⇒300 s

a = (v - u) / t = (20 m / s) / 300 s = 0.0667 m / s²

S = ut + 1 / 2 at² = 1 / 2 at²(Since u = 0) = (1 / 2)×(0.0667 m / s²)×(300 s)²

⇒(1 / 2)×(0.0667 m / s²)×(900 s²) = (1/2) × 6000 m = 3000 m = 3 km

Distance = 3 km

Acceleration = 0.0667 m / s

I hope it helps!