Question

A train is travelling at a speed of 90kmPh. Brakes are applied so as
to produce a uniform acceleration of -0.5ms-2. Find the distance travelled
by the train before it comes to rest.

Answer 1

Answer:

Given :-

• A train is travelling at a speed of 90 km/h.
• Brakes are applied so as to produce a uniform acceleration of - 0.5 m/s².

To Find :-

• What is the distance travelled by the train before it comes to rest.

Formula Used :-

$\clubsuit$ Third Equation Of Motion Formula :

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

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance Covered

Solution :-

First, we have to convert the initial velocity km/h into m/s :

$\implies \sf Initial\: Velocity =\: 90\: km/h$

$\implies \sf Initial\: Velocity =\: 90 \times \dfrac{5}{18}\: m/s\: \: \bigg\lgroup \sf\bold{\pink{1\: km/h =\: \dfrac{5}{18}\: m/s}}\bigg\rgroup$

$\implies \sf Initial\: Velocity =\: \dfrac{450}{18}\: m/s$

$\implies \sf\bold{\purple{Initial\: Velocity =\: 25\: m/s}}$

Now, we have to find the distance travelled by the train before it comes to rest :

Given :

• Final Velocity (v) = 0 m/s
• Initial Velocity (u) = 25 m/s
• Acceleration (a) = - 0.5 m/s²

According to the question by using the formula we get,

$\longrightarrow \sf (0)^2 =\: (25)^2 + 2(- 0.5) \times s$

$\longrightarrow \sf 0 \times 0 =\: 25 \times 25 + 2 \times \{- 0.5\} \times s$

$\longrightarrow \sf 0 =\: 625 + (- 1) \times s$

$\longrightarrow \sf 0 =\: 625 - 1s$

$\longrightarrow \sf 0 - 625 =\: - 1s$

$\longrightarrow \sf {\cancel{-}} 625 =\: {\cancel{-}} 1s$

$\longrightarrow \sf 625 =\: 1s$

$\longrightarrow \sf \dfrac{625}{1} =\: s$

$\longrightarrow \sf 625 =\: s$

$\longrightarrow \sf\bold{\red{s =\: 625\: m}}$

${\footnotesize{\bold{\underline{\therefore\: The\: distance\: travelled\: by\: the\: train\: before\: it\: comes\: to\: rest\: is\: 625\: m\: .}}}}$

Answer 2

⇝Given :-

• Speed of train = 90km/ph
• Acceleration = -0.5 m/s-²

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⇝To Find :-

• Distance covered = ?

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⇝Solution :-

❒We know that :

3rd equation of motion :

${\red{\bigstar \: \: \: \: {\green{\underbrace{\underline{\orange{\bf{V² = U² + 2as}}}}}}}}$

❒Here :

• U = Initial velocity = 90km/ph
• V = Final Velocity = 0m/s
• A = Acceleration = -0.5 m/s-²
• S = Distance covered = ?

❒Converting :

Speed :

${\mapsto{\sf{\cancel{90} \times \frac{5}{\cancel{18} }}}}$

${\blue{\dashrightarrow{\sf{\underline{25 m/s}}}}}$

Acceleration :

${\blue{\dashrightarrow{\sf{\underline{-0.5 m/s-²}}}}}$

❒Solving starts :

${\mapsto{\bf{V² = U² + 2as}}}$

${\twoheadrightarrow{\sf{(0)² = (25)² + 2 \times (-0.5) \times s}}}$

${\twoheadrightarrow{\sf{0 = 625 + (-1) \times s}}}</u></p><p><u>[tex]{\twoheadrightarrow{\sf{0 = 625 + (-1) \times s}}}$

${\twoheadrightarrow{\sf{0 - 625 = -1s}}}$

${\twoheadrightarrow{\sf{ {\cancel-} 625 ={\cancel -}1s}}}$

${\twoheadrightarrow{\sf{S = \frac{625}{1} }}}$

${\green{➳{\orange{\boxed{\red{\bf{S = 625m}}}}}}}$

❒Hence :

${\green{\orange{\underline{\boxed{\mathfrak{Distance Covered = 625 m}}}}}}$

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⇝More Info :

$\begin{gathered}\red{\large \qquad \boxed{\boxed{\begin{array}{cc} \ ➳ \: \: \bf v = u + at \\ \\ \ ➳ \: \: \bf s = ut + \dfrac{1}{2}a {t}^{2} \\ \\ \ ➳ \: \: \bf{v}^{2} - {u}^{2} = 2as\end{array}}}}\end{gathered}$