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?

Answer 1

Let $A$ represent the Old Steamy train and $B$ the Chug-a-Lug train. 
First, we are gong to find the distance from the Old Steamy train and the train station:
We know that the speed of the Old Steamy train is 6.14 mph. We also know that there are 2.5 hours from 8:00 am to 10:30 am; so using the distance equation:
$distance=velocity*time$
$d=vt$
$d=6.14mph*2.5hours$
$d=15.35miles$

Now, we know that the Old Steamy and Chug-a-Lug,are 290 miles apart from each other, so the distance between Chug-a-Lug and the station will be the distance from the Old Steamy to the station plus the distance between the trains:
$d_{B}=15.35mi+290mi$
$d_{B}=305.35miles$
Now that we have our distance, we are going to use the speed equation to find the speed of the Chug-a-Lug train:
$speed= \frac{distance}{time}$
$v= \frac{d}{t}$
$v= \frac{305.35miles}{2.5hours}$
$v=122.14mph$

We can conclude that the Chug-a-Lug must be going at 122.14 mph.

Answer 2

Answer:

122.14 mph.

Step-by-step explanation:

Given : 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.

To Find: 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?

Solution :

Speed of Old Steamy train = 6.14 mph

Train started at 8 a.m. and arrived at 10.30 a.m.

So, time of travel = 2.5 hours

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

                                                                      = $6.14 \times 2.5$

                                                                      = $15.35miles$

Since the total distance traveled by train Old Steamy is 15. 35 miles. This means Old steamy was standing before the Chug-a-Lug i.e. Chug-a-Lug was 290 miles behind the Old steamy

Distance traveled by Chug-a-Lug  is distance traveled by Old Steamy to the station plus the distance between the trains

Now, Distance traveled by Chug-a-Lug = 15.35+290 miles

                                                                 = 305.35

Since we are given that they started and ended at the same time . So, the the time of travel for both the trains will be same = 2.5 hours

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

                                            =$\frac{305.35}{2.5}$

                                            =$122.14$

Hence the speed of the Chug-a-Lug train is 122.14 mph.