Question

A train travels 120 km at a speed of 60 km/h, makes a stop for 0.5 h, and then travels the next 180 km at a speed of 90 km/h. What is the average speed of the train for this trip?

Answer 1

The average speed of the train for the trip is equal to $66.67\frac{km}{h}$

Why?

To calculate the average speed, we need to divide the trip into three moments:

- First moment, the train traveled 120 km at a speed of 60 km/h.

- Second moment, the train stopped for 0.5 hours, meaning that the train did not travel for that time (traveled distance equal to 0 km).

- Third moment, the train traveled 180 km at a speed of 90 km/h.

We can use the following formula to calculate the average speed:

$AverageSpeed=\frac{TotalDistance}{TotalTime}$

So, for our problem we have:

$AverageSpeed=\frac{distance_{1}+distance_{2}+distance{3}}{time_{1}+time_{2}+time_{3}}$

From the statement, we know that:

$distance_{1}=120km\\speed_{1}=60\frac{km}{h}\\\\distance_{2}=0km\\time_{2}=0.5hours\\\\distance_{3}=180km\\speed_{1}=90\frac{km}{h}$

Calculating the times, we have:

$speed=\frac{distance}{time}\\\\time=\frac{distance}{speed}$

$time_{1}=\frac{distance_{1}}{speed_{1}}=\frac{120km}{60\frac{km}{h}}=2hours$

$time_{3}=\frac{distance_{3}}{speed_{3}}=\frac{180km}{90\frac{km}{h}}=2hours$

Then, substituting the times into the main equation, we have:

$AverageSpeed=\frac{120km+0km+180km}{2hours+0.5hours+2hours}=\frac{300km}{4.5hours}=66.67\frac{km}{h}$

Hence, we have that the average speed of the train for the trip is equal to $66.67\frac{km}{h}$

Have a nice day!