Answer:
45 miles per hour = 66 feet/second
24 miles per hour = 35.2 feet/second
Train A will travel 3260.86 feet before they meet
Train B will travel 1739.13 feet before they meet
The train will meet in 49.41 seconds
Explanation:
Here is the complete question:
Two trains are 5000 feet apart, heading toward each other, but on separate parallel straight tracks, Train A is traveling at 45 miles per hour and Train B is traveling at 24 miles per hour.
Change 45 miles per hour and 24 miles per hour into feet per second.
About how far will each train travel before they meet? Round your answers to the nearest hundredth.
In how many seconds will the trains meet?
Explanation:
To change from miles per hour to feet per second, we will convert miles to feet and hour to second.
(NOTE: 1 mile = 5280 feet and 1 hour = 3600 seconds)
Hence,
For 45 miles per hour
45 miles per hour = 45 × 5280 feet / 3600 second
45 miles per hour = 66 feet/second
For 24 miles per hour
24 miles per hour = 24 × 5280 feet / 3600 second
24 miles per hour = 35.2 feet/second
To determine how far each train will travel before they meet,
Let the distance covered by Train A be x₁
the distance covered by Train B be x₂
and the time they meet be t
From
Speed = Distance / Time
Distance = Speed × Time
For Train A
Speed = 66 feet/second
∴ x₁ = 66t
For Train B
Speed = 35.2 feet/second
∴ x₂ = 35.2t
From the question, the two trains are 5000 feet apart.
Since the trains are 5000 feet apart, this means both trains will cover a total distance of 5000 feet.
∴x₁ + x₂ = 5000 feet
Then,
66t + 35.2t = 5000
101.2t = 5000
t = 5000/101.2
t = 49.407 seconds
t ≅ 49.41 seconds
But
x₁ = 66t
x₁ = 66 × 49.407
x₁ = 3260.86 feet
Hence, Train A will travel 3260.86 feet before they meet
For x₂
x₂ = 35.2t
x₂ = 35.2 × 49.407
x₂ = 1739.13 feet
Hence, Train B will travel 1739.13 feet before they meet
t is the time they meet
Hence, The train will meet in 49.41 seconds