Question

A train is travelling at a speed of 90 kmh-1
brakes are applied so as to produce a
uniform acceleration of -0.5m/s2
. Find how far the train will go before it is
brought to rest.

Answer 1

Answer:

$\displaystyle\huge\red{\underline{\underline{ANSWER}}}$

GIVEN:

u=90Km/h

u=90×1000/3600

u=25m/s

a=-0.5m/s²

S=?

v=0m/s

NOW:

v²-u²=2as

0²-(25)²= 2×-0.5×s

-s=-625

✈︎✈︎S=625m

Hence, train will go 625m

✍︎✍︎✍︎✍︎More equations of motion

V=u+at

v²-u²=2as

s=ut+1/2at²

Answer 2

Answer :-

625m

Explanation :-

Given :

Initial velocity of the train,u = 90km/h = 25m/s         [1km/h = 1000/(1*60*60)]

Final velocity of the train,v = 0m/s.   [As train is going to stop]

Acceleration,a = -0.5m/s^2

To Find :

Distance,s = ?

Solution :

According to third equation of motion:

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

Put thier values and solve for “s”

$\sf{}\implies 0=25^2m/s+2\times (-0.5s)\times s$

$\sf{}\implies 0=625-1.0s$

$\sf{}\implies -625=-1s$

$\sf{}\implies \dfrac{-625}{-1}=s$

$\sf{}\therefore s=625m$

Therefore,train will stop after 625m.

☆Know more☆

$\sf{}First\ equation\ of\ motion: v= u+at$

$\sf{}Second\ equation\ of\ motion: s=ut+\dfrac{1}{2}at^2$