math help?? please?!?

Question

Bubba leaves Eastside High School at 10 am walking east at 2 km/hr. At 10:30 am Bubbagirl leaves from the same place and walks south at 4 km/hr. At what time are they 5 km apart?

Work please?

Answer 1

You are making a right triangle here. Draw a picture -- it will help you visualize this the easiest.

at 11 am ... Bubba would be 2 km east and Bubbagirl would be 2km south. That triangle would have a hypotenuse of 2sqrt2 -- using pythagoreans theorem.

at 11:30am ... Bubba would be 3 km east and Bubbagirl would be 4 km south. Using pythogreans theorem again .. you would find out that they are 5 km apart.

The math is as follows ...

3² + 4² = ?²
9 + 16 = ?²
25 = ?²
5 = ?

Answer 2

At 10:30 Bubba is 1km away. Then Bubbagirl starts going at a right angle at twice the speed of Bubba. At 11:00 Bubba is 2 km away and so is Bubbagirl. This would be a 45 degree angle to both paths and Bubba squared plus Bubbagirl squared is 8 and the distance between is the square root of 8.
The distance is between them is the Hypotenuse which is equal to the square root of the sum of the two squares of the distance traveled by Bubba and Bubbagirl, or Bubba squared plus Bubbagirl squared equals the distance between them squared:
B^2 + BG^2 = D^2
This holds true no matter how far they go. What you are looking for is the time when D^2 = 25 (5 squared).
At 12:00 B^2 = 16 and BG^2 = 36 and is too far as 16 +36 = more than 25.
So at 11:00 D^2 = 8 and at 12:00 D^2 = 52
At 11:30 B^2 = 9 and BG^2 = 16 and D^2 = 25, so your answer is 11:30.

Answer 3

at 10:30, Bubba went 1 km.
the direction:
--------------------> Bubba
|
|
|
|
Bubbagirl

so the distance of these satisfies Pythagorean.
call t the time that Bubbagirl need to go.
so the length that Bubba went: 2t + 1
the length that Bubbagirl went: 4t
we have equation:
( 2t + 1 ) ^2 + 16t^2 = 25
=> 20t^2 + 4t -24 = 0
=> t = 1
so they are 5 km apart at 11:30

Answer 4

11:30

Answer 5

12:00