Question 158836This question is from textbook geometry CPM
: Don't really know what kind of problem this is. I used the examples from Question 157162 and Question 156861 on your website and of course came up with 2 different answers. I tried to get this done early for extra credit and can't get a consistent answer. Now my work is due tomorrow and I can't figure out the right answer. On a softball field, one person walks from home plate to 3rd base at 6' per sec. and one person jogs from home plate to 1st base at 10' per sec. How many seconds will it take them to be 270' apart? I submitted 16.875 sec. to my teacher and he said it was wrong because it has to with triangles and sqrt.
This question is from textbook geometry CPM
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
(1) Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Your teacher is right:
The softball diamond is a square. There's a 90 degree angle between home and first and home and third. When you draw a straight line between the two walkers/joggers, this straight line forms the hypotneuse of the right triangle.
What we want to know is how many seconds will have elapsed until this hypotneuse is 270' long.
Let t=time that elapses (seconds) until the persons are 270' apart:
Distance walker travels=6t-------one leg of the right triangle
Distance jogger travels=10t-------other leg of the right triangle
Now we know that (6t)^2+(10t)^2=(270)^2 or
36t^2 +100t^2=72,900 divide each term by 4
9t^2 +25t^2=18225 collect like terms
34t^2=18225 divide each side by 34
t^2=536.03 take sqrt of both sides
t=23.152 sec
CK
6*23.152=138.9 ft
10*23.152=231.52 ft
(138.9)^2+(231.52)^2=(270)^2
19293.21 + 53601.51=72,900
~~~~~~72,900=72,900---WE HAVE SOME ROUNDOFF ERROR
Hope this helps---ptaylor
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