Question

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

Answer 1

Question :-

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

Given :-

• A train is travelling at a speed of 90 km h-1. Brakes are applied so as to produce a uniform acceleration of -0.5m s-2

To find :-

• how far the train will go before brought the rest

formula :-

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

solution :-

now, first we need to convert the unit of intial speed from km/hr to m/s

$\rm :\longmapsto\: 90kmhr^{ - 1} = 90 \times \frac{5}{18} = 25ms^{ - 1}$

now by using formula

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

now by putting values in formula we get,

$\rm :\longmapsto\: (0)^{2} = (25)^{2} + 2( - 5)s$

$\rm :\longmapsto\: 0 = 625 + 2( - 5)s$

$\rm :\longmapsto\: 0 = 625 - 10s$

$\rm :\longmapsto\: 10s = 625$

$\rm :\longmapsto\: s = \cancel \frac{625}{10}$

$\rm :\longmapsto\: s = 62.5m$

so therefore the train will go before brought of rest by 62.5m

Answer 2

Answer:

Question :-

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

Given :-

A train is travelling at a speed of 90 km h-1. Brakes are applied so as to produce a uniform acceleration of -0.5m s-2

To find :-

how far the train will go before brought the rest

formula :-

\begin{gathered}\mapsto \sf\boxed{\bold{\pink{v^{2} = u^{2} + 2as }}}\\\end{gathered}

↦

v

2

=u

2

+2as

solution :-

now, first we need to convert the unit of intial speed from km/hr to m/s

\begin{gathered}\rm :\longmapsto\: 90kmhr^{ - 1} = 90 \times \frac{5}{18} = 25ms^{ - 1} \\ \end{gathered}

:⟼90kmhr

−1

=90×

18

5

=25ms

−1

now by using formula

\begin{gathered}\mapsto \sf\boxed{\bold{\pink{v^{2} = u^{2} + 2as }}}\\\end{gathered}

↦

v

2

=u

2

+2as

now by putting values in formula we get,

\begin{gathered}\rm :\longmapsto\: (0)^{2} = (25)^{2} + 2( - 5)s \\ \\ \end{gathered}

:⟼(0)

2

=(25)

2

+2(−5)s

\begin{gathered}\rm :\longmapsto\: 0 = 625 + 2( - 5)s \\ \\ \end{gathered}

:⟼0=625+2(−5)s

\begin{gathered}\rm :\longmapsto\: 0 = 625 - 10s \\ \\ \end{gathered}

:⟼0=625−10s

\begin{gathered}\rm :\longmapsto\: 10s = 625 \\ \\ \end{gathered}

:⟼10s=625

\begin{gathered}\rm :\longmapsto\: s = \cancel \frac{625}{10} \\ \\ \end{gathered}

:⟼s=

10

625

\begin{gathered}\rm :\longmapsto\: s = 62.5m \\ \\ \end{gathered}

:⟼s=62.5m

so therefore the train will go before brought of rest by 62.5m