Thread: Function Problem

Here is the problem:

Arthur is going for a run. From his starting point, he runs due east at 10 feet per second for 230 feet. He then turns, and runs north at 12 feet per second for 400 feet. He then turns, and runs west at 7 feet per second for 70 feet.

Express the (straight-line) distance from Arthur to his starting point as a function of t, the number of seconds since he started.



I can't figure out what I am doing wrong for the third part? Thank you!

Hey UWM120.

So for this problem, I am going to work out the hypotenuse of the triangle where the sides are the distance of the x and y positions.

So we start off with a east distance of 230 but we need to subtract the distance based on the fact that the runner is now travelling west. So the difference squared for the x distance (i.e. find the distance and square it for pythagoras) is (230 - 7t)^2.

Now for the y-distance, we know the guy is only running west so the y-distance doesn't change. The y-distance when the runner started running is 400 so the square is 400^2.

Now the hypotenuse is just the square root of the sum of squares which is

= SQRT(400^2 - (230 - 7t)^2)

Originally Posted by chiro

Hey UWM120.

So for this problem, I am going to work out the hypotenuse of the triangle where the sides are the distance of the x and y positions.

So we start off with a east distance of 230 but we need to subtract the distance based on the fact that the runner is now travelling west. So the difference squared for the x distance (i.e. find the distance and square it for pythagoras) is (230 - 7t)^2.



Now for the y-distance, we know the guy is only running west so the y-distance doesn't change. The y-distance when the runner started running is 400 so the square is 400^2.

Now the hypotenuse is just the square root of the sum of squares which is

= SQRT(400^2 - (230 - 7t)^2)

Ohhh okay I see now. Thank you very much!