Word Problem :: Two people at different times & rates

[kgh]

Hello!

This is my first post here, so please forgive my ignorance. Here is the word problem directly from the book:

Henry leaves Tankerville Square at 10AM walking east at 2km/h. At 10:30AM, Marlene leaves from the same place and walks south at 4km/h. At what time are they 5km apart?

We are currently solving problems by using monomials (specifically trinomials) and factoring (and setting them to equal 0, etc...).

So far... I have figured that it must be a right triangle, with the hypotenuse being 5km, and the two sides *possibly* being d=rt, so the two legs would be 4(x-30) and 2x because they would have to be rt, and one left 30 minutes after the other.

This gives me the following equation:
4x^2 + (4x-120)^2 = 25

When I factor this out, I get large numbers, which I am unsure if they are able to be factored using the general factoring method (__+/-__)(__+/-__).

If someone could please let me know if I am heading in the right direction, or if not, point me into a new direction.

 

[Denis]

K = time in hours by Henry; then Marlene's time = k - 1/2; so distances are:
Henry: 2k
Marlene : 4(k - 1/2) = 4k - 2

(2k)^2 + (4k - 2)^2 = 25
Solve for k; OK?

 

[soroban]

Re: Word Problem :: Two people at different times & rate

Hello, kgh!

Welcome aboard!

Henry leaves Tankerville Square at 10 AM walking east at 2 km/h.
At 10:30 AM, Marlene leaves from the same place and walks south at 4 km/h.
At what time are they 5km apart?

My approach is slightly different from Denis (who is absolutely correct).

Let $\displaystyle t$ = number of hours Marlene walked.
Henry had a half-hour headstart, so he walked $\displaystyle t\,+\,\frac{1}{2}$ hours.

Using $\displaystyle d\,=\,rt$:

Marlene walked for $\displaystyle t$ hours at 4 km/hr.
. . She walked $\displaystyle 4t$ km south.
Henry walked for $\displaystyle t+\frac{1}{2}$ hours at 2 km/hr.
. . He walked $\displaystyle 2\left(t\,+\,\frac{1}{2}\right) \:=\:2t\,+\,1$ km east.

When are they 5 km apart?
The situtation looks like this:

         2k+1
    T * - - - * H
      |      /
      |     /
   4t |    /
      |   / 5
      |  /
      | /
      |/
    M *

Using Pythagorus, we have: $\displaystyle \,(4t)^2\,+\,(2t\,+\,1)^2\;=\;5^2$

Sinplify: $\displaystyle \,16t^2\,+\,4t^2\,+\,4t\,+\,1\:=\:25\;\;\Rightarrow\;\;20t^2\,+\,4t\,-\,24\:=\:0$

Divide by 4: $\displaystyle \,5t^2\,+\,t\,-\,6\:=\:0$

This factors: $\displaystyle \,(t\,-\,1)(5t\,+\,6)\:=\:0$

Then: $\displaystyle \,\begin{array}{cc}t\,-\,1\:-\:0\;\;\Rightarrow\;\;t\,=\,1 \\ 5t\,+\,6\:=\:0\;\;\Rightarrow\;\;t\,=\,-\frac{6}{5}\end{array}$

We discard the "negative time" . . . so $\displaystyle t = 1$ hour.

Hence: Marlene walked for 1 hour, starting at 10:30 AM.


Therefore, there were 5 km apart at 11:30 AM.