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.

Answer 1

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.