Question

A train is moving at a speed of 54 km/h and acceleration of -4m/s2 . Find it's distance covered if driver applied breaks and train stops in 1 minute of time.​

Answer 1

Provided that:

• Initial velocity = 54 km/h
• Acceleration = -4 m/s²
• Time = 1 minute

↪️ Here, the acceleration is in negative means the train is moving in opposite direction of the motion, it is retarding.

To calculate:

• The distance if break are applied.

Solution:

• The distance = 13950 m

Using concept:

• Second equation of motion

• Formula to convert km/h into m/s

• Formula to convert min into s

Using formula:

• Second equation of motion is given by

• ${\small{\underline{\boxed{\sf{s \: = ut \: + \dfrac{1}{2} \: at^2}}}}}$

• Now minute into seconds,

• ${\small{\underline{\boxed{\sf{1 \: minute \: = 60 \: seconds}}}}}$

Required solution:

~ Firstly let us convert km/h into m/s!

$:\implies \sf 54 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{54} \times \dfrac{5}{\cancel{{18}}} \\ \\ :\implies \sf 3 \times 5 \\ \\ :\implies \sf 15 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Therefore, the initial velocity becames 15 metre per second!

~ Now let's convert min into seconds!

$:\implies \sf 1 \: minute \: = 60 \: seconds \\ \\ \sf It \: is \: already \: converted! \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Therefore, the time is 60 seconds!

~ Now let's use second equation of motion to calculate the distance!

$:\implies \sf s \: = ut \: + \dfrac{1}{2} \: at^2 \\ \\ :\implies \sf s \: = 15(60) + \dfrac{1}{2} \times (-4)(60)^{2} \\ \\ :\implies \sf s \: = 15(60) + \dfrac{1}{2} \times (-4)(3600) \\ \\ :\implies \sf s \: = 15(60) + \dfrac{1}{2} \times (-14400) \\ \\ :\implies \sf s \: = 900 + \dfrac{1}{2} \times (-14400) \\ \\ :\implies \sf s \: = \cancel{900} + \dfrac{1}{\cancel{{2}}} \times (-14400) \\ \\ :\implies \sf s \: = 450 + 1 \times (-14400) \\ \\ :\implies \sf s \: = 450 + (-14400) \\ \\ :\implies \sf s \: = 450 - 14400 \\ \\ :\implies \sf s \: = -13950 \\ \\ :\implies \sf Distance \: = +13950 \: m$

Additional information:

$\begin{gathered}\boxed{\begin{array}{c|cc}\bf Distance&\bf Displacement\\\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}&\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}\\\sf Path \: of \: length \: from \: which &\sf The \: shortest \: distance \: between \\ \sf \: object \: is \: travelling \: called \: distance. &\sf \: the \: initial \: point \: \& \: final \\ &\sf point \: is \: called \: displacement. \\\\\sf It \: is \: scalar \: quantity. &\sf It \: is \: vector \: quantity \\\\\sf It \: is \: positive \: always &\sf It \: can \: be \: \pm \: \& \: 0 \: too \end{array}}\end{gathered}$

Answer 2

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|cc}\bf Distance&\bf Displacement\\\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}&\frac{\qquad \qquad \qquad\qquad\qquad \qquad\qquad\qquad}{}\\\sf Path \: of \: length \: from \: which &\sf The \: shortest \: distance \: between \\ \sf \: object \: is \: travelling \: called \: distance. &\sf \: the \: initial \: point \: \& \: final \\ &\sf point \: is \: called \: displacement. \\\\\sf It \: is \: scalar \: quantity. &\sf It \: is \: vector \: quantity \\\\\sf It \: is \: positive \: always &\sf It \: can \: be \: \pm \: \& \: 0 \: too \\ & \sf Displacement ,\\ & \\ \sf Distance\end{array}}\end{gathered}\end{gathered}$