The speed of a passenger train is 6 mph faster than the speed of a freight train. The passenger train travels 280 miles in the same time it takes the freight train to travel 250 miles. Find the speed of each train.
What is the speed of passenger train? ____mph
What is the speed of freight train? _______ Mph
10) A long distance trucker travels 88 miles in one direction during a snowstorm. The return trip in rainy weather was accomplished at double the speed and took 2 hours less time. Find the speed going
The speed going was ____ mph.
P = speed of passenger train.
F = speed of freight train.
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P+6 = F
distance = rate x time or rearrange to
time = d/r
For P train time = 280/P
For F frain time = 250/F
The times are equal so
280/P = 250/F but you know P = F + 6 so substitute F + 6 for P to obtain
280/(F+6) = 250/F
solve for F.
Speed forward direction = F
Speed return trip = R
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R = 2F
The time it takes for forward is
t = d/r = 88/F
The time for return is
t = d/r = 88/R or 88/2F
IF the times were equal (they aren't) then
88/F = 88/2F BUT it takes the return trip 2 hours less; therefore, we must add 2 hours to the return trip to make the times equal.
88/F = (88/2F) + 2
solve for F, the speed in the forward direction.
To solve the first problem, we can use the formula: time = distance/speed.
Let's assume the speed of the freight train as 'x' mph.
According to the problem, the speed of the passenger train is 6 mph faster, so the speed of the passenger train will be 'x + 6' mph.
We know that the passenger train travels 280 miles in the same time it takes the freight train to travel 250 miles.
Using the formula, we can set up the equation:
280/(x + 6) = 250/x
We can cross-multiply to simplify the equation:
280x = 250(x + 6)
280x = 250x + 1500
Now, we can solve for x by subtracting 250x from both sides of the equation:
30x = 1500
Finally, we can solve for x by dividing both sides by 30:
x = 50
So, the speed of the freight train is 50 mph.
To find the speed of the passenger train, we can substitute the value of x back into the equation:
Speed of passenger train = x + 6 = 50 + 6 = 56 mph.
Therefore, the speed of the passenger train is 56 mph, and the speed of the freight train is 50 mph.
Now, moving on to the second problem.
Let's assume the speed of the truck while going as 'x' mph.
According to the problem, the return trip was accomplished at double the speed, so the speed of the truck while returning will be '2x' mph.
We know that the truck travels 88 miles in one direction, and the return trip took 2 hours less time.
Using the formula time = distance/speed, we can set up the equation:
88/x - 88/(2x) = 2
To solve this equation, we can first find a common denominator:
(88 * 2 - 88)/(2x) = 2
(176 - 88)/(2x) = 2
88/(2x) = 2
Now, we can cross-multiply to simplify the equation:
88 = 4x
Dividing both sides of the equation by 4, we get:
x = 22
So, the speed going is 22 mph.
Therefore, the speed going was 22 mph.