Question

A train travelling at a speed of 75kmph. the brakes are applied so as to produce a uniform acceleration of -0.5m/s².find how far the train goes before it stops​

Answer 1

$\huge \sf {\orange {\underline {\pink{\underline{Solution :-}}}}}$

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 =?

${\pink{\boxed{\large{\bold{Formula's}}}}}$

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

★We have to find the distance covered before it stops .

→By Equation of motion

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

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

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

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

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

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

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

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

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

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