Question

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. Find how for the train will go before it is brought to rest.

Answer 1

Answer:

Given:

Initial Velocity = 90 km/hr

Acceleration = -0.5 m/s²

To find:

Distance travelled before after which the train stopped.

Concept:

Since the acceleration is constant , we can easily apply the equations of Kinematics and find out the required Displacement.

Conversion:

We need to convert unit of speed from km/hr to m/s.

90 km/hr = 90 × (5/18) = 25 m/s

Calculation:

As per Equation of Kinematics , we can say :

${v}^{2} = {u}^{2} + (2 \times a \times s)$

$= > {0}^{2} = {(25)}^{2} + 2 \times ( - 0.5) \times s$

$= > s = {(25)}^{2}$

$= > s = 625 \: m$

So final answer :

$\boxed{ \red{ \huge{ \bold{s = 625 \: m}}}}$

Answer 2

$\red{\color{white}{\fcolorbox{cyan}{black}{Answer:-}}}$

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Speed of the train(v) = 90 km/h = 25 m/s

Acceleration (a) = -0.5 m/s²

Final velocity (v)= 0 ...(because the train will come to rest

Initial velocity (u) = 90

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We know that:-

$Acceleration = \frac{v - u}{t} \\ {\implies}- 0.5 = \frac{0 - 25}{t} \\{\implies} - \frac{1}{2} = \frac{ - 25}{t} \\{\implies} t = ( - 25)( - \frac{2}{1} ) \\{\implies} t \: = \: 50$

∴ Time (t) = 50seconds

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We know that:-

$s \: = ut + \frac{1}{2} a {t}^{2}$

Where s = distance traveled

Here,

s=distance traveled by the train before it is brought to rest.

${\implies}s \: = 25 \times 50 + \frac{1}{2} \times ( - 0.5) \times (50) {}^{2} \\ {\implies} s \: = 25 \times 50 + \frac{1}{2} \times ( - \frac{1}{2} ) \times 50 \times 50 \\{\implies} s \: = 1250 + 25 \times ( - 25) \\{\implies} s \: = 1250 - 625 \\{\implies} s \: = 625$

∴ Distance (s) = 625m

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Your answer:-

$\huge{\boxed{\red{s=625m}}}$

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