Question

Between two stations a train accelerates from the rest uniformly at first, then moves with constant velocity, and finally retards uniformly to come to rest. If the ratio of the time taken is $8:1$ and maximum speed attained is $60\,km/h$ then what is the average speed over the whole journey?

Answer 1

Here we will first know about the equations of motion, how many and which are they. Then by using those equations in a proper manner we will solve the given problem.The average speed is the total distance traveled by the object in a particular time interval. The average speed is a scalar quantity. It is represented by the magnitude and does not have direction.

Formula used:
Equation of motion:
First equation of motion:
$v=u+at$
Second equation of motion:
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
Third equation of motion:
${{v}^{2}}={{u}^{2}}+2as$
Where $t$ is the time taken by the object to travel the given distance, $s$ is the displacement, $u$ is the initial velocity, $v$ is the final velocity, and $a$ is the acceleration.

Complete step by step answer:
The equations which relate quantities like velocity, time, acceleration and displacement provided the acceleration remains constant are collectively known as the equation of motion. There are three equations of motion. Here we will use each of them to solve the given problem.

Here we have, speed =$60\,km/hr$
So by using first equation of motion,
$60=v=u+a{{t}_{1}}=a{{t}_{1}}$ (because train starts from the rest)
$\Rightarrow a=\dfrac{60}{{{t}_{1}}}$
Also, by second equation of motion:
${{s}_{1}}=u{{t}_{1}}+\dfrac{1}{2}a{{t}_{1}}^{2}=\dfrac{1}{2}\dfrac{60}{{{t}_{1}}}{{t}_{1}}^{2}=30{{t}_{1}}$
Now, by third equation of motion:
${{v}^{2}}={{u}^{2}}+2a{{s}_{1}}$
From ${{s}_{1}}=\dfrac{1}{2}a{{t}_{1}}^{2}$
We substitute
$a=\dfrac{2{{s}_{1}}}{{{t}_{1}}^{2}} \\\ \Rightarrow v=\dfrac{2{{s}_{1}}}{{{t}_{1}}}$
The distance is travelled by the train at this velocity. Therefore we get,
${{s}_{2}}=v{{t}_{2}}=\dfrac{2{{s}_{1}}}{{{t}_{1}}}{{t}_{2}}$
But we know that the ratio of time is given as:
$\dfrac{{{t}_{2}}}{{{t}_{1}}}=\dfrac{8}{1}$
${{t}_{2}}=8{{t}_{1}} \\\ \Rightarrow {{s}_{2}}=16{{s}_{1}}$
And here we got
${{s}_{1}}=30{{t}_{1}} \\\ \Rightarrow {{s}_{2}}=480{{t}_{1}}$
We are given that ${{s}_{3}}={{s}_{1}}=30{{t}_{1}}$
Therefore we get the average speed as:
$\dfrac{Distance\text{ }travelled}{\text{ }time\text{ }taken}=\dfrac{30{{t}_{1}}+480{{t}_{1}}+30{{t}_{1}}}{{{t}_{1}}+8{{t}_{1}}+{{t}_{1}}} \therefore \dfrac{Distance\text{ }travelled}{\text{ }time\text{ }taken} =54\,km/hr$

Therefore the average speed of the train over the whole journey will be $54\,km/hr$.

Note: One should remember the equations of motion in order to be able to solve such kind of motion related (kinematic) problems. We just have to carefully identify what all parameters are given and then choose the appropriate equation and solve for the required parameter. The equations of motion are also used in the determination of optical properties.