Question

A train travelling at a speed of 75kmph. the brakes are applied so as to produce a uniform acceleration of -0.5m/s^2.find how far the train goes before it stops

Answer 1

Answer:

434

Explanation:

Given

u = 75km/hr = 20.83m/s²

a = -0.5 m/s²

v= 0m/s²

S=?

so by using equation of motion

v²-u²=2as

s=(0²-20.83²)/(2(-0.5))

s=434m

So, the train will go 432m more

Answer 2

Given :

Initial velocity, u = 75 km/h = 20.83 m/s²

Final velocity, v = 0

Uniform acceleration, a = -0.5 m/s²

To Find :

Distance covered, s =?

tex]{\purple{\boxed{\large{\bold{Formula's}}}}}[/tex]

Kinematic equations for uniformly accelerated motion .

$\bf\:v=u+at$

$\bf\:s=ut+\frac{1}{2}at{}^{2}$

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

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$

Solution :

We have to find the distance covered before it stops .

By Equation of motion

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

$\sf\implies\:0=(20.83)^{2}+2(-0.5)\times\:S$

$\sf\implies\:(20.83)^{2}=2\times0.5\times\:S$

$\sf\implies\:433.88=2\times0.5\times\:S$

$\sf\implies\:S=\dfrac{433.88}{2\times0.5}$

$\sf\implies\:S=\dfrac{43388\times10}{2\times5\times100}$

$\sf\implies\:S=\dfrac{43388}{2\times5\times10}$

$\sf\implies\:S=\dfrac{43388}{100}$

$\bf\implies\:S=433.88m$