How would you create an equation and solve for the following problem?San Francisco and Los Angeles are 470 miles apart by train. An express train leaves Los Angeles at the same time a passenger...
How would you create an equation and solve for the following problem?
San Francisco and Los Angeles are 470 miles apart by train. An express train leaves Los Angeles at the same time a passenger train leaves San Francisco. The express train travels 10 miles per hour faster than the passenger train. The 2 trains pass each other in 2.5 hours. How fast is each train traveling?
Assume that the distance between SF and LA = 470
The passanger train speed = P
the Express train speed = E = P+10
after 2.5 hours, Express train traveled D miles from LA
Passanger train traveled 470-D from SF
we need to find the E and P
we know that Speed = distance/time
so P = (470-D)/2.5
E = D/2.5
but E = P + 10
then p+10 = D/2.5
and P = (470-D)/2.5
By substituting in the first equation,
(470-D)/2.5 + 10= D/2.5
(470-D + 25 = D
==> 495 = 2D
==> D = 495/2 = 247.5
the P = (470-D)/2-5 = (470-247.5)/2.5= 89 m/h
and E = P+10 = 99 m/h
For America, these are very fast trains! The passenger train is going 89 mph and the express is going 99 mph. Here is how to figure that out.
Let x be the speed of the express and p be the speed of the passenger train.
x = p + 10
The speeds they go multiplied by 2.5 must equal 470 because the two of them between them have traveled 470 miles when they meet. So
2.5x + 2.5p = 470
2.5 (p+ 10) + 2.5p = 470
2.5p = 25 + 2.5p = 470
5p = 445
p = 89
Let the speed of the passenger and express train be x and x+20 mph.
The distance travelled by the trains in 2.5 hours are;
2.5(x+x+20) which is 470 m together when the meet.
so 2x+20 = 470/2.5 =188. )r
2x = 188-10 = 178. Or
x = 178/2 = 89 is the speed of the ordinary train and 89+10 = 99 mph is the speed of the express train