Question

A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of −0.5m/s2. Find how far will the train go before it is brought to rest.

Answer 1

Given:

• Speed of traing = 90 km/h
• After applying brakes deceleration produced = -0.5 m/s²

To find :

• Distance travelled after applying brakes

Solution:

Using the 2nd equation of motion

v² - u² = 2as

Where

• v : final velocity
• u : initial velocity
• a : acceleration or deceleration
• s : distance travelled

Here , initial velocity = 90km/h = 25m/s & final velocity = 0 m/s because the train is applied brakes and comes to rest

→ 0² - 25² = 2(-0.5)(s)

→ 0 - 625 = 2 (-½) ( s )

→ - 625 = - s

→ S = 625m

Hence , distance travelled after applying brakes = 625 m

Answer 2

Given :-

Initial velocity of the train, u = 90km/hr

Final speed of the train, v = 0

Acceleration, a = -0.5 m/s^2

To Find :-

Distance acquired.

Formula to be used :-

Here we using third equation of motion:

➾ v^2 = u^2 + 2as

Here,

• v = Final velocity
• u = Initial velocity
• a = acceleration
• s = distance

Solution :-

Putting all the values in the equation, we get

➾ v^2 = u^2 + 2as

➾ (0)^2 = (25)^2 + 2 (-0.5) s

➾ s = 625 m

Hence train will cover a distance 625 m before it is brought to rest.