Question

two trains,Old Steamy and Chug-a-Lug,are 290 miles apart from each other and headed for the same station. They started toward the station at 8:00 a.m. If they are both set to arrive at 10:30 a.m. and Old Steamy is going 6.14 mph, how fast must Chug-a-Lug be going? How far does Chug-a-Lug have to travel?

Answer 1

8 am to 10:30 am is 2.50 hours.

Old Steamy is going 6.14 mph for 2.5 hours so travels : 6.14 * 2.5 = 15.35 miles.

This means Chug a lug has to travel 290 - 15.35 = 274.65 miles in 2.5 hours.

The speed would be 274.65 / 2.5 = 109.86 miles per hour.

Answer 2

Answer:

The speed of Old Steamy train = 6.14 mph

Train started at 8 a.m. and arrived at 10.30 a.m, this becomes 2.5 hours.

So, distance traveled by Old Steamy train = $Speed\times time$

=> $6.14\times2.5=15.35$ miles

Now we get that Chug-a-Lug was 290 miles behind the Old steamy.

To know the distance traveled by Chug-a-Lug, we will add distance traveled by Old Steamy to the station plus the distance between the trains.

So, distance traveled by Chug-a-Lug = $15.35+290=305.35$ miles

The speed of Chug-a-Lug =$\frac{Distance}{Time}$

= $\frac{305.35}{2.5}=122.14$ mph