Question

A train accelerating uniformly from rest attains a maximum speed of 40m/s in 20sec. It travels at this speed for 20sec and is brought to rest with uniform retardation in 40 sec. The average velocity during this period is
$\begin{aligned} & \text{a)}\left( \dfrac{\text{80}}{\text{3}} \right)\text{ m/s} \\\ & \text{b)30 m/s} \\\ & \text{c)25 m/s} \\\ & \text{d)40 m/s} \\\ \end{aligned}$

Answer 1

It is given to us how the train travels along its distance from its initial velocity zero to its final velocity which also equals zero. The average velocity is defined as the total distance covered in the total interval of time. Hence we can determine the distance traveled by the train when it accelerates and decelerates at different intervals of time and divides its total time of the journey. We will use Newton’s first and second kinematic equation to determine the total distance covered.

Complete answer:
Newton’s first kinematic equation is given as,
$V=U+at...(1)$ where V is the final velocity of the body, U is the initial velocity of the body a is the magnitude of the acceleration of the body and t is the time for which the body is under acceleration a.
Similarly if the body travels with the same parameters as mentioned above in time t, then the displacement of the body is given by,
$S=Ut+\dfrac{1}{2}a{{t}^{2}}...(2)$.
Initially it is given to us that the train accelerates uniformly for 20 sec from rest and reaches a maximum speed of 40m/s. Hence its acceleration from the equation 1 we get,

$\begin{aligned} & V=U+at \\\ & 40=0+a(20) \\\ & a=2m{{s}^{-2}} \\\ \end{aligned}$

Hence the distance covered by the train during this period is equal to,

$\begin{aligned} & S=Ut+\dfrac{1}{2}a{{t}^{2}} \\\ & S=0(20)+\dfrac{1}{2}2{{\left( 20 \right)}^{2}} \\\ & S=400m \\\ \end{aligned}$

Further the train travels for 20 sec with its maximum speed. Hence the distance covered for 20sec is,
$\begin{aligned} & S=Ut+\dfrac{1}{2}a{{t}^{2}} \\\ & S=40(20)+\dfrac{1}{2}0{{\left( 20 \right)}^{2}} \\\ & S=800m \\\ \end{aligned}$

Finally the train comes to rest from this point with uniform retardation in 40 sec. hence from the deceleration of the train from equation 1 is,
$\begin{aligned} & V=U+at \\\ & 0=40+a(40) \\\ & a=-1m{{s}^{-2}} \\\ \end{aligned}$
Hence the distance covered by the train during this period is,
$\begin{aligned} & S=Ut+\dfrac{1}{2}a{{t}^{2}} \\\ & S=40(40)+\dfrac{1}{2}-1{{\left( 40 \right)}^{2}} \\\ & S=1600-800 \\\ & S=800m \\\ \end{aligned}$
Hence the total distance of the entire journey of the train is the sum of all the distances i.e. equal to 2000m and the total time taken for the journey to get completed is 80 sec’s. Since the average speed is the ratio of the total distance covered to the total time taken for the journey to get completed, the average speed of the train is equal to,
$\begin{aligned} & \text{Average speed=}\dfrac{\text{total distance}}{\text{Total time}} \\\ & \text{Average speed=}\dfrac{2000m}{80\sec } \\\ & \text{Average speed=}25m/s \\\ \end{aligned}$

So, the correct answer is “Option C”.

Note:
In the above case the train accelerates either uniformly or decelerates uniformly. Hence the above kinematic equations are valid. If the train was under non uniform acceleration, we would have to integrate for distance with respect to the non uniform acceleration.