9. A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce uniform retardation of
0.5 m/s2. Find how far the train will go before it is brought to rest.
Answers
Given :
• Initial velocity of the train = 90 km/h
• Final velocity of the train = 0 m/s
• Retardation of the train = 0.5 m/s²
To find :
• Distance the train will travel before coming to rest
Solution :
Here, we shall use the third equation of motion to find the distance the train will travel before coming to rest. But before that we will make the units same for the right answer.
To convert the value of initial velocity from km/h to m/s mulitply it by 5/18.
→ Initial velocity = 90 × 5/18
→ Initial velocity = 5 × 5
→ Initial velocity = 25
Therefore, initial velocity of the train = 25 m/s
Now, calculating the distance :-
→ Third equation of motion :-
- v² - u² = 2as
where,
- v = Final velocity
- u = Initial velocity
- a = Acceleration
- s = distance/displacement
We have,
- v = 0 m/s
- u = 25 m/s
- a = - 0.5 m/s² [Here we have taken the value of acceleration as negative because in the question we are given the value of retardation and negative acceleration is known as retardation.]
Substituting the given values :-
→ (0)² - (25)² = 2(0.5)(s)
→ - 625 = - 1 × s
→ 625 = s
Therefore, the train will travel 625 m before coming to rest.