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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Acacia 3 years 2021-08-25T01:21:44+00:00 1 Answers 12 views 0

Answers ( )

    0
    2021-08-25T01:23:31+00:00

    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.

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