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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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 =?

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

Formula

′

s

Kinematic equations for uniformly accelerated motion .

\bf\:v=u+atv=u+at

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

2

1

at

2

\bf\:v{}^{2}=u{}^{2}+2asv

2

=u

2

+2as

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

nth

=u+

2

a

(2n−1)

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

→By Equation of motion

\rm\:v{}^{2}=u{}^{2}+2asv

2

=u

2

+2as

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

2

+2(−0.5)×S

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

2

=2×0.5×S

\sf\implies\:433.88=2\times0.5\times\:S⟹433.88=2×0.5×S

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

2×0.5

433.88

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

2×5×100

43388×10

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

2×5×10

43388

\sf\implies\:S=\dfrac{43388}{100}⟹S=

100

43388

\bf\implies\:S=433.88m⟹S=433.88m

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