A train left Podunk and traveled north at 75 Km/h. Two hours later, another train left Podunk and traveled in the same direction at 100 km/h. How many hours had the first train traveled when the second train overtook it?
distance = rate x time
A train rate = 75 km/hr
B train rate = 100 km/hr
distance traveled by A train =
75 x (t+2 hrs)
distance traveled by B train =
100 x t
Distance is the same so set them equal to each other.
75(t+2) = 100t
solve for t = time.
You can solve it algebraically or you can set up a chart:
First train:
1 hr: 75 mi
2 hr: 150 mi
3 hr: 225 mi
-- Continue this chart
Second train:
1 hr: 100 mi
2 hr: 200 mi
3 hr: 300 mi
-- Continue this chart.
Compare the charts to see when they've both traveled the same distance, but the second train reached it in two hours less.
i needhelp with solving systems of equetions algebrically
Use the matrix method to solve:
x + 4y = 8
2x + y = 9
x = a0
y = a1
To find out how many hours the first train had traveled when the second train overtakes it, we need to set up an equation.
Let's assume the second train overtakes the first train after t hours.
The first train has been traveling for t + 2 hours because it left two hours earlier.
Distance traveled by the first train = speed × time = 75 × (t + 2)
Distance traveled by the second train = speed × time = 100 × t
When the second train overtakes the first train, both trains would have traveled the same distance. Therefore, we can set up the equation:
75 × (t + 2) = 100 × t
Now, let's solve the equation to find the value of t.
75t + 150 = 100t (applying distributive property)
150 = 100t - 75t (combined like terms)
150 = 25t
t = 150 / 25
t = 6
So, the second train overtakes the first train after 6 hours.
To answer the original question, the first train had traveled for 6 + 2 = 8 hours when the second train overtook it.