Question

A train is travelling at a speed of
90 km h-1 Brakes are applied so
as to produce a uniform
acceleration of - 0.5 m s-2. Find
how far the train will go before it
is brought to rest.​

Answer 1

Given :-

• Initial velocity of train = 90km/h

• Final velocity= 0 ( break applied )

• Acceleration (retardation)= -0.5m/s^2

To find :-

• Distance covered by the train before brought in rest .

Using third equation of motion

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

v= final velocity

u= initial velocity

a= acceleration

s= distance

Convert initial speed of car into m/s

$\bullet\sf 1 \ km/h = \dfrac{5}{18}\ m/s\\ \\ \implies\sf \cancel{90}\times \dfrac{5}{\cancel{18}}= 5\times 5\\ \\ \implies\sf u= 25m/s\\ \\ \implies\sf v= 0$

Now find distance

$\implies\sf (0)^2= (25)^2+2\times (-0.5)\times s\\ \\ \implies\sf 0= 625 -1s\\ \\ \implies\sf -625= -1s\\ \\ \implies\sf \cancel{\dfrac{-625}{-1}}= s\\ \\ \implies\sf 625= s$

$\underline{\bigstar{\sf\ \ Distance \ covered \ by \ train = 625m}}$

Answer 2

${\bold{\pink{\underline{\green{G}\purple{iv}\orange{en}\red{:-}}}}}$

$\bf\:Initial\:Speed\:(u)=90\:km/h$

$\bf\:Initial\:Speed\:(u)=\dfrac{5}{18}\times\:90$

• $\bf\:Initial\:speed\:(u)=25\:m/s$

• $\bf\:Final\:speed\:(v)=0\:m/s$

• $\bf\:acceleration=-0.5\:m/{s}^{2}$

${\bold{\blue{\underline{\red{To}\:\pink{Fi}\green{nd}\purple{:-}}}}}$

• $\bf\:Required\:distance=?$

${\bold{\pink{\underline{\red{So}\purple{lut}\green{ion}\orange{:-}}}}}$

$\bf\boxed\star\purple{\underline{\underline{{Using\:third\:equation\:of\:motion:-}}}}$

we know that,

$\pink{\boxed{\bf{{v}^{2}={u}^{2}+2as}}}$

$\bf\:{0}^{2}={25}^{2}+2\times(-0.5)\times\:s$

$\bf\:s=625\:m$

Hence, Required distance will be 625 m.

Other equations of motion:-

$\green{\boxed{\bf{v=u+at}}}$

$\pink{\boxed{\bf{s=ut+\dfrac{1}{2}a{t}^{2}}}}$