Question

On a 60km track, a train travels the first 30km with a speed of $30km/h$. How fast must the train travel the next 30km so as to average $40km/h$ for the whole trip?

Answer 1

Step 1: Given data

1. The speed of the train in the first half is $v_1=30km/h.$.
2. The distance of the track is $S=60km.$
3. The average speed of the train is $V_{avg}=40km/h$.

Step 2: Formulae and concept

1. We know that speed is the time rate of change in distance, i.e, $speed=\frac{dis\tan ce}{time}$.
2. The average velocity of a body is $V_{avg}=\frac{totaldis\tan cetraveledbythebody}{totaltimetaken}$.

Step 3: Finding the speed at second the half

Let $v_2$ be the velocity of the train in the second half, and the total time traveled by train in the second half is $t_2$ seconds. And $t_1$ be the time taken in the first half with a speed $v_1=30km/h$.

Now,

Again, as we know, $time=\frac{dis\tan ce}{speed}$.

So, the total time is taken in the first half is,

$$\begin{gathered} t_1=\frac{30}{30}=1 \\ or{t}_1=1hour.............\left(1\right) \end{gathered}$$

Similarly, the total time is taken in the second half is,

$t_2=\frac{30km}{v_2}................\left(2\right)$

Now, average speed, $V_{avg}=40km/h$

$$\begin{gathered} V_{avg}=\frac{totaldis\tan cetraveledbythebody}{totaltimetaken}=\frac{60}{t_1+t_2}=\frac{60}{1+{\displaystyle \frac{30}{v_2}}} \\ or40=\frac{60v_2}{v_2+30} \\ or40v_2+1200=60v_2 \\ or60v_2-40v=1200 \\ or20v_2=1200 \\ or{v}_2=\frac{1200}{20}=60 \\ or{v}_2=60km/h \end{gathered}$$

Now, velocity of the train in second half is $60km/h.$