Question

A train is scheduled to run from Delhi to Pune at an average speed of $80\text{km/h}$ but due to repairs of track , it looses $2\text{hrs}$ in the first part of the journey. It then accelerates at a rate of $20{\text{km/h}}^2$ till the speed reaches $100\text{km/h}$. This speed is now maintained till the end of the journey. If the train now reaches station on time, then find the distance from pune to a point when it starts accelerating.

• A. $840\text{km}$
• B. $640\text{km}$
• C. $550\text{km}$
• D. Data insufficient

Answer 1

The correct option is A $840\text{km}$

Given,
Average speed $=80\text{km/hr}$


Let $l$ be the total distance between Delhi and Pune. From the figure,

Scheduled time , $t_{exp}=\frac{l}{80}$

For travelling $x_1$ distance it looses $2\text{hrs}$ from expected time. So,

$t_1=\frac{x_1}{80}+2......\left(1\right)$

For travelling $x_2$ it goes under acceleration from $80\text{km/h}$ to $100\text{km/h}$.

So, applying equation of motion for $x_2$ distance,

$t_2$=$\frac{v-u}{a}=\frac{100-80}{20}=1\text{hr}$

$x_2=u{t}_2+\frac{1}{2}a{t}_2^2=80\times 1+\frac{1}{2}\times 20\times 1^2=90\text{km}$

For travelling $(x_3=l-x_1-x_2)$ it continues with $100\text{km/h}$,

$t_3=\frac{x_3}{100}=\frac{l-x_1-x_2}{100}$

Finally it reaches on time. So,

$t_{exp}=t_1+t_2+t_3$

Now, substituting the values,

$\Rightarrow \frac{l}{80}=(\frac{x_1}{80}+2)+1+\left(\frac{l-x_1-x_2}{100}\right)$

Substituting the value of $x_2$ as $90\text{km}$,

$\frac{l}{80}=\frac{x_1}{80}+3+\frac{l-x_1-90}{100}$

$\Rightarrow \frac{l}{80}-\frac{x_1}{80}=3+\frac{l-x_1-90}{100}$

$\Rightarrow (l-x_1)(\frac{1}{80}-\frac{1}{100})=3-\frac{90}{100}$

$\Rightarrow (l-x_1)\left(\frac{20}{80}\right)=210$

$\Rightarrow l-x_1=840\text{km}$

Distance of point $\text{A}$ from Pune $=l-x_1$

$=840\text{km}$

Hence, option (a) is the correct answer.