Question

A train stationery position and moving with uniform acceleration attains a speed of 36 km/ h in 10 min . Find its acceleration.

Answer 1

Answer:

216 km/h²

Explanation:

As we know, a = $\frac{v}{t}$

Given :

v = 36 kmph

t = 10 min

Procedure :

(a) t = 10 min, But velocity is given in km/h

Hence converting t from minutes to hours gives, t = 10/60 hrs

∴ t = 1/6 hrs

Now a = v/t

⇒ a = $\frac{36}{(1/6)}$

⇒ a = 36 × 6

⇒ a = 216 km/hr²

(OR)

(b) t = 10 min

Converting t in minutes to seconds, t = 10 × 60 = 600 s

Converting v in km/h into m/s = 36 × (5/18) m/s

⇒ v = 10 m/s

∴ a = 10/600 m/s²

∴ a = 1/60 m/s²

Thanks !