Question

A train is travelling at a speed of 50kmph when the driver decided to apply brakes with a

retardation speed of 18m/s2

. Find the time required to stop the train.​

Answer 1

Given : A train is travelling at a speed of 50kmph when the driver decided to apply brakes with a retardation speed of 18m/s²

To Find  : the time required to stop the train

Solution:

Train initial Speed = 50 km/hr

1 km = 1000 m

1hr = 3600 Secs

Hence speed in m /s  = 50 * 1000 / 3600

= 500 / 36

= 250 / 18  m/s

a = - 18 m/s²

v = 0   at rest

v = u + at

=> 0  = 250/18  + (-18) t

=> t = 250/18²

=> t = 0.7716 sec

time required to stop the train.​  = 0.7716 sec

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Answer 2

Given:

A train is travelling at a speed of 50kmph when the driver decided to apply brakes with a

retardation speed of 18m/s².

To find:

Time required to stop train?

Calculation:

First of all, convert unit of velocity from km/hr to m/s :

$50 \: kmph = 50 \times \dfrac{5}{18} = 13.88 \: m {s}^{ - 1}$

Now, applying 1st Equation of Kinematics:

$\therefore \: v = u + at$

$\implies 0 = 13.88 + ( - 18)t$

$\implies 0 = 13.88 - 18t$

$\implies 18t = 13.88$

$\implies t \approx 0.77 \: sec$

So, time taken to stop is approximately 0.77 sec.