SOLUTION: A freight train leaves the town of Harfville at 1:00PM travelling due to east at constant speed. Marcial, a Hobo, sneaks onto the train and fall asleep. AT the same time, Chokie le

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: A freight train leaves the town of Harfville at 1:00PM travelling due to east at constant speed. Marcial, a Hobo, sneaks onto the train and fall asleep. AT the same time, Chokie le      Log On

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Question 887106: A freight train leaves the town of Harfville at 1:00PM travelling due to east at constant speed. Marcial, a Hobo, sneaks onto the train and fall asleep. AT the same time, Chokie leaves Harfville on her bicycle, travelling along a straight road in a northeasterly direction (but not due northeast)at 10 miles per hour. At 1:12PM, Marcial rolls over in his sleep and falls from the train onto the side of the tracks. He wakes up and immediately begins walking at 3.5 miles per hour directly towards the road on which Chokie is riding. Marcial reaches the road at 2:12PM, just as Chokie is riding by. What is the speed of the train in miles per hour? Answer and solution please.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Both start at 1.00 pm
let train speed be x mph
Marcial travels by train for 12 minutes= 1/5 hours
distance traveled by Marcial = (1/5) * x = x/5 miles
Then Marcial walks for 1 hour perpendicular to the train.
speed = 3.5 mph
time = 1 hour
distance = 3.5 miles
Chokie rides 1 hour + 12 minutes to reach Marcial.
=6/5 hours
chokie's speed = 10 mph
distance chokie meets Marcial = 6/5 * 10 = 12 miles
So apply Pythagoras theorem
12^2= (x/5)^2+ (3.5)^2
144-12.25 = x^2/25
131.75 =x^2/25
x^2= 3293.75
x=57.4
Train speed = 57.4 mph