Question

A train is travelling at a speed of 72km/h.The driver applies brakes so that a uniform acceleration of 0.2m/s^2is produced.Find the distance travelled by the train before it comes to stop.

Answer 1

Answer:

Explanation:

Solution,

Here, we have

Initial velocity, u = 72 km/h = = 72 × 5/18 = 20 m/s

Final velocity, v = 0 (As brakes applied)

Acceleration, a = -0.2 m/s²

To Find,

Distance traveled s = ?

According to the 3rd equation of motion,

We know that,

v² - u² = 2as

So, putting all the values, we get

⇒ v² - u² = 2as

⇒ (0)² - (20)² = 2 × (- 0.2) × s

⇒ 400 = - 0.4s

⇒ 400/- 0.4 = s

⇒ s = 1000 m or 1 km.

Hence, the distance traveled by train is 1000 m or 1 km.

Answer 2

Answer:

1000 m

Explanation:

(hoping this is a straight line motion, where speed = magnitude of velocity)

u (initial velocity) = 72km/h =$72 *\frac{1000}{60*60} = 72*\frac{1000}{3600} = 20m/s$

v (final velocity) = 0m/s (rest)

a = -0.2 m/s^2

by 3rd equation of motion

2as= v^2-u^2

2* -0.2 s = 0^2 - 20^2

2* -0.2 s = 0 -400 = -400

-0.2s = $\frac{-400}{2}$ = -200

s = $\frac{-200}{-0.2} = \frac{-2000}{-2}$

= 1000 m

Hope this helps

pls.... mark as brainliest