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Ray starts walking to school at a rate of 2 mi/h. Ten minutes later, his sister runs after him with his lunch, averaging 6 mi/h.
Question
Ray starts walking to school at a rate of 2 mi/h. Ten minutes later, his
sister runs after him with his lunch, averaging 6 mi/h. Write a system of
linear equations to represent this situation.
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Mathematics
3 years
2021-08-25T01:21:44+00:00
2021-08-25T01:21:44+00:00 1 Answers
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Answer:
We know that Ray walks to school at a rate of 2mi/h, then the equation that represents the distance to his house is:
R(t) = (2mi/h)*t
Where t represents time in hours.
And we know that his sister has a velocity of 6mi/h, but she starts 10 mins after Ray.
We want to write 10 minutes in hours.
1 hour = 60 minutes.
1 = (1/60) hours/minute.
Then 10 minutes = 10 minutes*(1/60 hours/minute) = (10/60) hours = 0.167 hours.
Then the equation that represents the distance to her house can be written as:
S(t) = 6mi/h*(t – 0.167h)
where the “-0.167h” part is because she starts 0.167 hours after her brother.
Then the system of equations is:
R(t) = (2mi/h)*t
S(t) = 6mi/h*(t – 0.167h)
Such that the solution of this system is when both of them are at the same distance from their house at the same time, this happens when:
R(t) = S(t)
Replacing thee equations we get:
(2mi/h)*t = 6mi/h*(t – 0.167h)
Now we want to solve this for t.
(6mi/h)*0.167h = (6mi/h)*t – (2mi/h)*t = (4mi/h)*t
((6mi/h)*0.167h)/ (4mi/h) = t = 0.2505 h
This means that her syster reachs him 0.2505 hours after he leaves his house.