Question

A train starting from rest attains a velocity of $72\,kmh{r^{ - 1}}$ in $5\,\min $. Assuming the acceleration is uniform. Find the acceleration and distance travelled by train for this velocity.

Answer 1

Hint
The acceleration of the train is determined by using the acceleration equation of motion and by using the velocities given in the question. And the distance travelled by the terrain is determined by the second equation of the motion formula which gives the distance.
The acceleration equation of the motion is given by,
$\Rightarrow a = \dfrac{{v - u}}{t}$
Where, $a$ is the acceleration of the train, $v$ is the final velocity of the train, $u$ is the initial velocity of the train and $t$ is the time taken by the train.
The distance is given by the second equation of motion by,
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}$
Where, $s$ is the distance travelled by the train, $u$ is the initial velocity of the train, $t$ is the time taken by the train and $a$ is the acceleration of the train.

Complete step by step answer
Given that, The train starts from the rest that means the initial velocity is zero, $u = 0\,m{s^{ - 1}}$
The final speed of the train is, $v = 72\,kmh{r^{ - 1}}$
To convert the unit from $kmh{r^{ - 1}}$ to the $m{s^{ - 1}}$, the speed value is multiplied by $\Rightarrow \dfrac{5}{{18}}$, then the final velocity for the given speed is, $v = 20\,m{s^{ - 1}}$
The time taken by the train is, $t = 5\,\min = 300\,\sec $
Now, The acceleration equation of the motion is given by,
$\Rightarrow a = \dfrac{{v - u}}{t}\,..................\left( 1 \right)$
By substituting the initial velocity, final velocity and time taken in the above equation (1), then the above equation (1) is written as,
$\Rightarrow a = \dfrac{{20 - 0}}{{300}}$
The above equation is written as,
$\Rightarrow a = \dfrac{{20}}{{300}}$
On dividing the above equation, then the above equation is written as,
$\Rightarrow a = \dfrac{1}{{15}}\,m{s^{ - 2}}$
Now, The distance is given by the second equation of motion by,
$\Rightarrow s = ut + \dfrac{1}{2}a{t^2}\,.................\left( 2 \right)$
By substituting the initial velocity, final velocity, acceleration and time taken in the equation (2), then the equation (2) is written as,
$\Rightarrow s = \left( {0 \times 300} \right) + \dfrac{1}{2}\left( {\left( {\dfrac{1}{{15}}} \right) \times {{\left( {300} \right)}^2}} \right)$
By multiplying the terms, then the above equation is written as,
$\Rightarrow s = \dfrac{1}{2}\left( {\left( {\dfrac{1}{{15}}} \right) \times {{\left( {300} \right)}^2}} \right)$
By using the square, then the above equation is written as,
$\Rightarrow s = \dfrac{1}{2}\left( {\left( {\dfrac{1}{{15}}} \right) \times 90000} \right)$
By multiplying the terms inside the bracket, then
$\Rightarrow s = \dfrac{1}{2} \times \dfrac{{90000}}{{15}}$
On further simplification, then the above equation is written as,
$\Rightarrow s = \dfrac{{90000}}{{30}}$
On dividing the terms in the above equation, then
$\Rightarrow s = 3000\,m = 3\,km$.
The acceleration of the train is $a = \dfrac{1}{{15}}\,m{s^{ - 2}}$
The distance travelled by the train is $s = 3000\,m = 3\,km$

Note
The formulas which are used in this solution are the basic formulas in physics to find the acceleration and the distance. The initial velocity is taken as zero because in the question it is given that the train starts from rest. So, the initial velocity is zero.