Question

(ii) Brakes were applied on a train running at 60km/h to stop in 5 minutes on a coming railway
station. Calculate distance travelled by the train before it comes to rest.
(5)

Answer 1

Given:-

• Initial velocity ,u = 60km/h

• Time taken ,t = 5min

• Final velocity ,v = 0m/s

To Find:-

• Distance covered by the train ,s.

Solution:-

Conversion Of Unit

60km/h = 60×5/18 = 50/3m/s = 16.66m/s

Time = 5×60s = 300s

Firstly we need to calculate the acceleration of the train.

Using 1st equation of motion

• v = u +at

Substitute the value we get

→ 0 = 16.66 +a×300

→ -16.66 = a×300

→ a = -0.055 m/s²

Here negative sign show retardation

Now By Using 3rd equation of motion

→ v² = u² +2as

Substitute the value we get

→ 0² = 16.66² + 2×(-0.055)×s

→ -277.55 = -0.11×s

→ s = -277.55/-0.11

→ s = 2523.18m

∴ The distance covered by the train is 2.52km.

Answer 2

Answer:

Initial velocity ,u = 10m/s

Acceleration ,a = 8m/s²

Final velocity ,v = 40m/s

So By using 3rd equation of motion

• v² = u² +2as

Substitute the value we get

→ 40² = 10² +2×8×s

→ 1600 = 100 + 16×s

→ 1600-100 = 16×s

→ 1500 = 16×s

→ s = 1500/16

→ s = 93.75m

The length of Runway 93.80m