Question

A train is travelling at a speed of 90km/hr brakes are applied so as to produce a uniform acceleration - 0
5m/s2 find how far the train will go before it is brought to rest

Answer 1

Given:

A train is travelling at a speed of 90km/hr brakes are applied so as to produce a uniform acceleration of - 5m/s².

To find:

Distance travelled by train before it comes to rest.

Calculation:

First , we need to convert the unit of initial speed from km/hr to m/s :

$90 \: km {hr}^{ - 1} = 90 \times \dfrac{5}{18} = 25 \: m {s}^{ - 1}$

Now , applying 3rd equation of kinematics:

$\therefore \: {v}^{2} = {u}^{2} + 2as$

$= > \: {(0)}^{2} = {(25)}^{2} + 2( - 5)s$

$= > \: 0 = 625 + 2( - 5)s$

$= > \: 0 = 625 - 10s$

$= > \: 10 s= 625$

$= > \: s= \dfrac{625}{10}$

$= > \: s= 62.5 \: m$

So, final answer is:

Distance travelled before coming to rest is 62.5 m.

Answer 2

GivEn :-

A train is travelling at a speed of 90km/h, Brakes are applied so as to produce a uniform Acceleration of - 5 m/s 2

To FinD :-

• Distance travelled by train before it Comes to Rest.

SoluTioN :-

In the Question is to find the distance travelled by a Train of 90 km/h

Converting 90km/h to m/s by Multiplying 5/18

$\bold{ \pink{ {90kmh}^{ - 1} = 90 \times \frac{5}{18} = {25ms}^{ - 1} }{} }{}$

So The speed = 25m/s

By Applying lll Equation of Motion we get,

${v}^{2} = {u}^{2} + 2as \\ \\ {0}^{2} = {(25)}^{2} + 2( - 5)s \\ \\ 0 = 625 - 10s \\ \\ 10s = 625 \\ \\ s = \frac{625}{10} = 62.5meter$

Hence distance travelled by train before reaching to rest = 625m