Question

THE BAKES APPLIED TO A TRAIN MOVING AT 90 KM/H PRODUCES A RETARDATION OF 5 m/s2. WHAT DISTANCE WILL IT COVER BEFORE COMING TO A STOP? 6.25 M 625 M 62.5M

Answer 1

$\bigstar$ Answer $\bigstar$

$\checkmark$ Given:

Brakes applied to a train moving at 90 km/h produce a retardation of 5 m/s²

$\checkmark$ To Find:

Distance will it cover before coming to a stop

$\checkmark$ Solution:

We know that,

$\star$ Equation of Motion

$\red{\boxed{\sf v^{2}=u^{2}+2as}}$

Here,

s = Distance

u = Initial Velocity

v = Final Velocity

a = Acceleration

According to the Question,

We are asked to find the distance will it cover before coming to a stop

Hence,

We are asked to find Distance (s)

Given that,

u = 90 km/h

Converting in to m/s

u = 25 m/s

v = 0 m/s (coming to stop)

a = - 5 m/s² (retardation)

Substituting the above values,

We get,

$\blue{\sf \implies 0^{2}=25^{2}+2(-5) \times s}$

$\green{\sf \implies 0=625-10s}$

$\orange{\sf \implies 10s=625}$

$\red{\sf \implies s=\dfrac{625}{10} \ m}$

$\pink{\sf \implies s=62.5 \ m}$

Therefore,

$\orange{\boxed{\bold{\rm Distance \ (s)=62.5 \ m}}}$

Hence,

Train will cover 62.5 m before coming to stop

Therefore,

$\checkmark$ Option (3) is correct

Answer 2

✰ Given -

• Initial Velocity (u) = 90 km/h
• Final Velocity (v) = 0 m/s ..(Brakes r applied)
• Acceleration = 5 m/s^2

✰ To Find -

• Distance travelled by train

✰ Solution -

We know 2nd Equation Of Motion,

$✓ \: {v}^{2} = {u}^{2} + 2as$

Here,

• V = Initial Velocity
• U = Final Velocity
• A = Acceleration
• S = Distance

Here,

Converting km/h to m/s

$⟶ u = \frac{90}{3.6}$

$⟶u = \frac{\cancel{90}}{\cancel{3.6}}$

$⟶u = 25m</strong><strong>/</strong><strong>s</strong><strong>$

$⟶a = - 5 {m</strong><strong>/</strong><strong>s}^{2}$ ..(Deceleration)

Placing Values,

$⟶( {0}^{2}) = ( {25}^{2} ) + 2 \times ( - 5) \times s$

$⟶0 = 625 - 10s$

$⟶10s = 625 - 0$

$⟶0 = 625 - 10s$

$⟶10s = 625$

$⟶s = \frac{\cancel{625}}{\cancel{10}}$

$⟶s = 62.5m$

⛬ Total Distance covered is 62.5m