Word Problem: Find distance as a function of time.

[Chaim]

Bob goes to run. From his starting point, he runs due east at 10 ft per sec for 250 ft. He then turns and runs north at 12 ft per sec for 400 ft. He then turns and runs west at 9 ft per sec for 90 ft.
Express the (straight-line) distance from Bob to his starting point as a function of t, the number of seconds since he started.


This was the graph , or sketch I MADE about this problem, so if you have an easier idea, feel free to post, since it says he keeps turning to the direction in which he turn in
So the first part was d=s(t) from his starting point (Point A) to his second point (Point B), so we know that the speed there was 10 ft/s so then d=10t
Now I have to find the others

I know that the equation we would be using is the distance formula, but since time is included, this is making it difficult for me. Can anyone help explain to me what do you do on this problem?

 

[tkhunny]

It WILL have to be piece-wise, right? Do the first 25 seconds and then we can talk.

 

[Chaim]

It WILL have to be piece-wise, right? Do the first 25 seconds and then we can talk.

I think I got it kind of
c = sqrt (a^2 + b^2)
So time = 25 seconds from line A to b since t = d/s = 250/10 = 25
So we know the first part from line AB = 10t
So AC = d=(250^2 + 12(t-25))
250 line AB
12 is the speed from BC
the t-25 is the time, and -25 is from subtracting the time from line AB

Am I right?

This is the second multipart function.

By the way, did you notice that there was over 1000 views, and yet I only got 1 reply so far? xD

 

[tkhunny]

c = sqrt (a^2 + b^2)

You had it here.

So AC = d=(250^2 + 12(t-25))

But somehting funny happened, here.

For t <= 25, you have \(\displaystyle s_{1}(t) = 10t\)

For 25 < t < 25+33+(1/3), you have \(\displaystyle s_{2}(t) = \sqrt{250^{2} + [12(t-25)]^{2}}\)

How about the last piece?

 

[Chaim]

You had it here.



But somehting funny happened, here.

For t <= 25, you have \(\displaystyle s_{1}(t) = 10t\)

For 25 < t < 25+33+(1/3), you have \(\displaystyle s_{2}(t) = \sqrt{250^{2} + [12(t-25)]^{2}}\)

How about the last piece?

Oopsies, forgot to do the square roots and squaring the speed and time together xD
So the last piece, since I know a^2 = 400^2
Then it's d=sqrt 400^2 + s(t)
Speed = 9 ft/sec
t = subtract from line AB and BC, so 400/12 = 100/3, and 25=75/3, so that means -175/3
So sqrt(400^2 + (250+9(t-175/3))^2)
I think, since the 250 is also part of b too.

This time varible thing has me troubling xD
Only learned it for a day.

 

[tkhunny]

90/9 = 10

250 - 9*(t-(25+33+(1/3)))