Question

A train is travelling at a speed
of 90 km hrl. 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 1

Given :

▪ Initial velocity = 90kmph

▪ Retardation = 0.5m/s^2

To Find :

▪ Distance covered by the train before it is brought to rest.

Solution :

→ This question is completely based on concept of stopping distance.

→ Since, acceleration is constant throughout the motion, we can easily apply equation of kinematics directly.

→ Third equation of kinematics is given by

☞ v^2 - u^2 = 2aS

✒ 1kmph = 5/18mps

✒ 90kmph = 90×5/18 = 25mps

• v = zero (i.e. rest)

→ (0)^2 - u^2 = 2(-a)S

Negative sign shows retardation

→ -u^2 = -2aS

→ S = u^2/2a

→ S = (25×25)/(2×0.5)

→ S = 625m

Answer 2

Given :

• Initial velocity (u) = 90 km/h
• Retardation (a) = - 0.5 m/s²
• Final velocity (v) = 0 m/s

To Find :

• How far will train go, if brakes are applied.

Solution :

We are given initial velocity is in km/h, We have to convert it in m/s. So, we have to multiply it by 5/18

$\implies \sf{Initial \: velocity \: = \: 90 \: \times \: \dfrac{5}{18} \: = \: 25 \: ms^{-1}}$

Now, A.T.Q,

$\underbrace{\sf{Distance \: travelled \: after \: breaks \: are \: applied}}$

Use 3rd equation of Kinematics

$\implies \sf{v^2 \: - \: u^2 \: = \: 2as} \\ \\ \implies \sf{0^2 \: - \: 25^2 \: = \: 2 \: \times \: -0.5 \: \times \: s} \\ \\ \implies \sf{-25^2 \: = \: -s} \\ \\ \implies \sf{s \: = \: 625}$

$\therefore$ Distance Traveled by train after applying breaks is 625 m

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Additional Information :

• Retardation means the negative acceleration, acceleration is negative when it is applied in opposite direction of motion.

• SI unit of Acceleration is m/s²

• It is a vector quantity