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(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=(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(km)/(h)\\\\distance_(2)=0km\\time_(2)=0.5hours\\\\distance_(3)=180km\\speed_(1)=90(km)/(h)\\`

Calculating the times, we have:

`speed=(distance)/(time)\\\\time=(distance)/(speed)`

`time_(1)=(distance_(1))/(speed_(1))=(120km)/(60(km)/(h))=2hours`

`time_(3)=(distance_(3))/(speed_(3))=(180km)/(90(km)/(h))=2hours`

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

`AverageSpeed=(120km+0km+180km)/(2hours+0.5hours+2hours)=(300km)/(4.5hours)=66.67(km)/(h)`

Hence, we have that the average speed of the train for the trip is equal to
`66.67(km)/(h)`

Have a nice day!