Question

A train starts from rest and accelerates uniformly at a rate of $2m{{s}^{-2}}$ for $10s$ . It then maintains a speed for $200s$ . The brakes are then applied and the train is uniformly retarded and comes to rest in $50s$ . Find the average velocity of the train.
(A). $35.4m{{s}^{-1}}$
(B). $20m{{s}^{-1}}$
(C). $17.69m{{s}^{-1}}$
(D). $8m{{s}^{-1}}$

Answer 1

Using equations of motions for different intervals in which acceleration is constant, calculate the distance travelled. Sum the distances to find total distance travelled. The average speed is the total distance travelled divided by total time taken.

Formula used:
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
$v=u+at$
$\text{average}\,\text{speed=}\dfrac{\text{total}\,\text{distance}\,\text{travelled}}{\text{total}\,\text{time}\,\text{taken}}$

Complete step by step solution:
The train starts from rest and accelerates at $2m{{s}^{-2}}$ for $10s$ . Using the following eq-
$s=ut+\dfrac{1}{2}a{{t}^{2}}$ - (1)
Here, $u$ is initial velocity
$v$ is final velocity
$s$ is distance travelled
$t$ is time taken
$a$ is acceleration
Substituting values In eq (1), we get,

& s=0\times 10+\dfrac{1}{2}\times 2\times {{(10)}^{2}} \\\ & s=100m \\\ \end{aligned}$$ Therefore, the train travels $$100m$$ in the first $$10s$$ . Next, using the following eq we calculate speed for next $$200s$$ - $$v=u+at$$ - (2) Substituting values in the above eq, we get, $$\begin{aligned} & v=0+2\times 10 \\\ & v=20m{{s}^{-1}} \\\ \end{aligned}$$ The train maintains a speed of $$20m{{s}^{-1}}$$ for the next $$200s$$ , the distance travelled in this time is- $$\begin{aligned} & \text{speed=}\dfrac{\text{distance}}{\text{time}} \\\ & 20=\dfrac{x}{200} \\\ & 20\times 200=x \\\ & x=4000m \\\ \end{aligned}$$ In $$200s$$ the distance travelled by the train is $$400m$$ . Next the train comes to rest in $$50s$$ . Using eq (2), the final acceleration will be- $$\begin{aligned} & 0=20-a\times 50 \\\ & a=\dfrac{2}{5}m{{s}^{-2}} \\\ \end{aligned}$$ Substituting values in eq (1), we calculate the distance travelled in $$50s$$ as- $$\begin{aligned} & s=20\times 50-\dfrac{1}{2}\times \dfrac{2}{5}{{(50)}^{2}} \\\ & s=500m \\\ \end{aligned}$$ $$$$ Therefore, the total distance travelled by the train is $$100+4000+500=4600m$$ . The total time taken by the train is $$10s+200s+50s=260s$$ . The formula for average speed is- $$\text{average}\,\text{speed=}\dfrac{\text{total}\,\text{distance}\,\text{travelled}}{\text{total}\,\text{time}\,\text{taken}}$$ Substituting values in the above eq- $$average\,speed=\dfrac{4600}{260}=17.69m{{s}^{-1}}$$ The average velocity of the train is $$17.69m{{s}^{-1}}$$ . **So, the correct answer is “Option (C)”.** **Note:** The equations of motion in a straight line can be used only when the body is moving with constant acceleration, this means no external forces are acting on it. According to the second law of motion, force is required to change the state of rest or motion of a body. Therefore, force is applied to bring the train from rest to motion; similarly, a braking force is applied to bring it back to rest.