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This is a problem I had in AP Physics today:

Two cars travel in the same direction along a straight
highway, one at a constant speed of 55 mi/h and the
other at 70 mi/h. (a) Assuming that they start at the
same point, how much sooner does the faster car arrive
at a destination 10 mi away? (b) How far must the faster
car travel to be before it a 15-min lead on the slower car?

Say f(x) = 55x and g(x) = 70x
Distance is the y-axis (in miles)
Time is the x-axis (in hours)
I want to find what g(x) equals when g(x)-f(x)=15
How do I do this on a calculator?
You want to find g(x) when g(x)-f(x)=15 on a calculator?
What type of calculator do you mean?

If it's a TI-84 or similar I think you could graph
Y1=55x
Y2=70x
Y3=Y2-Y1

and could either search through the table until you found the part where Y3=15, or you could do

Y4=15 and use the intersect option in the trace menu on Y3 and Y4 if I am understanding what you want correctly.

Then, when you find the intersect (when Y3 == 15) just plug in the corresponding X value to Y2 by entering Y2(xval)
Minxrod wrote:
You want to find g(x) when g(x)-f(x)=15 on a calculator?
What type of calculator do you mean?

If it's a TI-84 or similar I think you could graph
Y1=55x
Y2=70x
Y3=Y2-Y1

and could either search through the table until you found the part where Y3=15, or you could do

Y4=15 and use the intersect option in the trace menu on Y3 and Y4 if I am understanding what you want correctly.

Then, when you find the intersect (when Y3 == 15) just plug in the corresponding X value to Y2 by entering Y2(xval)

That worked. I already had the answers I just didn't know how to implement it on the calculator. The only problem is, 70 - 55 = 15 so Y3 actually hits 15 at 1 hour, or x = 1, when the actual answer is like 2 mins or so less than an hour. I don't know how to get around that problem.
Edit:
I just noticed that I actually needed when the lines were 15 minutes apart, or .25 apart, according to the x-axis.
A tortoise can run with a speed of 10.0 cm/s, and a hare can run 20 times as fast. In a race, they both start at the same time, but the hare stops to rest for 2.0 minutes. the tortoise wins by a shell (20 cm). (a) How long does the race take? (b) What is the length of the race?
B) 1220 cm long
A)2 min 2 sec

Because the hare stopped for 2 min. , and the tortoise is crawling at 10 cm/s, that works out to the 1200 cm long. Add 20 cm to that (the finishing distance), and that works out to 2 seconds, adding up to a total of 1220 cm and giving us 2 min 2 sec.

Voila!
In the time the hare is resting, the tortoise can run 120*10 = 1200 cm. The difference in speed of the hare and the tortoise is 190 cm/s. Because the tortoise wins with the difference of 20 cm, the hare needs to bridge 1180 cm. The difference was 190 cm/s so the time would be 2 min + 1180/190 = +- 126.21 sec. The hare now runned 6.21*200 = 1242 cm and the tortoise 126.21*10 = 126.21 which is about a difference of 20 cm.
a) 126.21 seconds
b) 1262.11 centimeter
PT_ is correct. I figured this out after a very long 2 hours but that's because I didn't read the question fully till about 20 minutes before that. Then I used the plan described above and it got me the answer. I did not know how to do it mathematically, that's why I asked. Hello, it is a 2 mobiles RUM.

First, write the 2 equations:

x_f = x_0 + v*t

first part:

10 = 0 + (55/60)*t => t= 10,9 min

Second part: Solve this sistem for t (I recommend you the Cramer method, is the method used in calculators)

x = 0 + (55/60)*t
x - 15 = 0 + (70/60)*t

I expect I helped!
frankmar98 wrote:
Hello, it is a 2 mobiles RUM.

First, write the 2 equations:

x_f = x_0 + v*t

first part:

10 = 0 + (55/60)*t => t= 10,9 min

Second part: Solve this sistem for t (I recommend you the Cramer method, is the method used in calculators)

x = 0 + (55/60)*t
x - 15 = 0 + (70/60)*t

I expect I helped!

You lost me there. I get what you're doing a bit however the 0+ is throwing me off because I don't know why it's there. I understand the 55/60. However, the way PT_ explained it is much easier to comprehend. Thanks though. 0 is x_0, the initial position, the distance between the initial position and the reference sistem.

You can put the reference sistem where is more simple, or in any other point.

I use the equation: final position = initial position + v*t,
you can extract it from v=x/t
Here are some physics equations that can be used in free-fall motion which is becoming common I guess in calculator games(maybe I'm just crazy I don't know):
v=v_o + at
x=x_o + v_ot + (1/2)at^2
v^2=v_o^2 + 2a(x - x_o)

v=final velocity
v_o=vknot or initial velocity
a=acceleration(Gravity is 9.8m/s^2 on earth)
t=time
x=final position
x_o=xknot or initial position
Yup, those are the kinematics equations. Which, I'm guessing, is what you were supposed to use for the above questions. Takes me back to junior year *shudders*
GTemples27 wrote:
Yup, those are the kinematics equations. Which, I'm guessing, is what you were supposed to use for the above questions. Takes me back to junior year *shudders*

Actually no I wasn't. I only got these equations 3 weeks ago. Those problems above I had 4-5 weeks ago. They might be able to be solved with those equations, but they were not meant to be.
Ah, okay. It's just that my teacher just dove right into the unit and was all equation-happy.
Can anyone explain to me Big G and how it relates to black holes?

But basically though it relates "Big G" to the escape velocity from a black hole. This is overly simplistic though; I would recommend reading more in depth seanlego23 wrote:
(1/2)at^2

Substitute r for t, and τ for a.
/me runs
Q: Is your physics class being taught using calculus techniques, or just algebra and memorization?
I'm guessing algebra; it doesn't seem like the calc-based physics course.

But then, I've never taken Physics C, so who knows? ¯\_(ツ)_/¯
Algebra physics. It's unfortunate. Physics C seems like it would be easier for me. As long as you get to use graphs and calculators to do the work instead of showing all of your work...but that's just me.

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