Question

A train is travelling at a speed of 90 km h-1. Brakes are applied so as to produce a uniform acceleration of ?0.5 m s-2. Find how far the train will go before it is brought to rest.

Answer 1

Answer:-

Given:-

$Initial \: speed \: of \: the \: train, u = 90 \: km/h$

$we \: will \: covert \: km/h \: to \: m/s$

$1 \: km/h$ = $\frac{5}{18} \: m/s$

$90 \: km/h$ = $90$ × $\frac{5}{18}$

Then, Initial speed of the train, u = 25 m/s

Final speed of the train, v = 0 (Because at last train comes at rest and its velocity will Zero)

Acceleration =   ₋ $0.5 \: m/s$⁻²

Solution:-

Here we will use third equation of motion:  $v^{2}$ = $u^{2}$ + $2as$

Where, s is the distance covered by the train.

⇒ $v^{2} = u^{2} + 2as$

⇒ $(0)^{2}= (25)^{2} + 2 ( - 0.5) s$

⇒ $s = \frac{25^{2} }{2(0.5)}$

⇒ $s = 625$

The train will cover a distance of 625 m before coming to rest.

Answer 2

$Question:$

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

$Answer:$

Train will travel 625 m before it is bought to rest.

$Explanation:$

According to the question the speed of the train is = 90 km/h

$= 90 \times \frac{5}{18} m {s}^{ - 1} \\ \\ = 5 \times 5 m {s}^{ - 1} \\ \\ = 25m {s}^{ - 1}$

Retardation of the train = -0.5 m/s²

Final velocity = 0 m/s

By using the equation we can find distance covered by train before it is bought to rest:

=> v² = u² + 2as

By putting the value we get

=> 0² = 25² + 2(-0.5)s

=> 0 = 625 - 1s

=> s = 625

Therefore, Train will travel 625 m before it is bought to rest.