Question

# A train starting from a railway station and moving with uniform acceleration attains a speed of $40km{{h}^{-1}}$ in 10 minutes. Find its acceleration in the units of $m{{s}^{-2}}$.\begin{align} & \text{A}\text{. }4 \\ & \text{B}\text{. }0.067 \\ & \text{C}\text{. }0.0185 \\ & \text{D}\text{. }1.11 \\ \end{align}

Hint: As the train starts from rest, we will take its initial velocity to be zero. We know 1 hour contains 3600 seconds and 1 kilometer has 1000 meters, so we will convert the given final speed of train from $km{{h}^{-1}}$ to $m{{s}^{-1}}$. We will also convert the given time of 10 minutes into seconds; using 1 minute has 60 seconds. Finally we will apply the first equation of uniformly accelerated motion i.e. $v=u+at$ to find the acceleration of the train in $m{{s}^{-2}}$.

Formula used: First equation of uniformly accelerated motion: $v=u+at$

We have, a train starting from a railway station and moving with uniform acceleration attains a speed of $40km{{h}^{-1}}$ in 10 minutes,
As the train is starting from the station, the initial velocity of the train will be zero.
Let, $u$ represents the initial velocity of the train,
So, $u=0$
Now, in 10 minutes, the train attains a speed of $40km{{h}^{-1}}$
As the required acceleration is in $m{{s}^{-2}},$
We will convert the given time from minutes to seconds,
And, given speed from $km{{h}^{-1}}$ to $m{{s}^{-1}}$
We know that in a minute there are 60 seconds,
So, 1 minute is equal to 60 seconds
And, 10 minutes will be equal to: $10\times 60$ seconds
Or, 10 minutes will be equal to: 600 seconds
Now, we know that in 1 kilometer, there are a thousand meters, and in an hour there are 3600 seconds.
So, $1km=1000m$ and $1hour=3600\text{seconds}$
Or, $1km{{h}^{-1}}=\dfrac{1km}{1hour}$
Or, $1km{{h}^{-1}}=\dfrac{1000m}{3600\text{seconds}}$
Or, $1km{{h}^{-1}}=\dfrac{5}{18}m{{s}^{-1}}$
Thus, we have $1km{{h}^{-1}}=\dfrac{5}{18}m{{s}^{-1}}$
So, $40km{{h}^{-1}}$ will be equal to:
\begin{align} & 40km{{h}^{-1}}=40\times \dfrac{5}{18}m{{s}^{-1}} \\ & 40km{{h}^{-1}}=\dfrac{200}{18}m{{s}^{-1}} \\ & 40km{{h}^{-1}}=\dfrac{100}{9}m{{s}^{-1}} \\ \end{align}
Thus, the train attains a speed of $\dfrac{100}{9}m{{s}^{-1}}$ in 600 seconds.
Now, we know the first equation of uniformly accelerated motion is:
$v=u+at$
Where,
$v$ is the final velocity of the object
$u$ is the initial velocity of the object
$a$ is the uniform acceleration of the object
$t$ is the time taken by the object to reach velocity $v$ with uniform acceleration $a$
Now, for the train, we have
Initial velocity, $u=0$
Final velocity, $v=\dfrac{100}{9}m{{s}^{-1}}$
Time taken, $t=600\sec$
So, substituting the given values in the equation $v=u+at$
We have,
\begin{align} & \dfrac{100}{9}m{{s}^{-1}}=0+a\times 600\sec \\ & \dfrac{100}{9}m{{s}^{-1}}=a\times 600\sec \\ & a=\dfrac{\dfrac{100}{9}m{{s}^{-1}}}{600\sec } \\ & a=\dfrac{100}{9\times 600}m{{s}^{-2}} \\ & a=\dfrac{1}{54}m{{s}^{-2}} \\ & a=0.0185m{{s}^{-2}} \\ \end{align}
The acceleration of the train is $0.0185m{{s}^{-2}}$

So, the correct answer is “Option C”.

Note: For uniformly accelerated motion, we can apply any of the equations of motion in accordance with the terms given to us. While solving these types of numericals, every term should be taken in the SI units only to avoid any kind of calculation error.