CaptainBlood

OK, thanks, I did it.

Two cars start moving form the same place. One goes north at a rate of 50 mi/hr, while the other heads east at a rate of 30 mi/hr. At what rate is the distance between the two cars changing exactly two hours later? I thought that the equation for the velocity vectors is: r(x) = 30xi + 50yj and thus distance d two hours later is given by: d = √(30x² + 50x² ) and dd/dt = ∂x/dt + ∂y/dt and after solving the equation I just plug in 2 for x and y to get what time the distance changes at the two hour mark. Is this right?

Ahoy, OK, thanks 4/5 I can't believe I didn't see that. Arrrhhhhhhh

If the product of two numbers if 5, and the sum of these same two numbers is 4, what's the sum of the reciprocals of these two numbers? I tried: let's say we have two numbers a and b, then a*b = 5 a = 5/b since their product is 4, we get b + 5/b = 4 or b^2 -4b + 5 = 0 but this has no real solutions just imaginary because this function is > 0 for all b I can't factor it and the quadratic formula doesn't work since 4^2 < 4*1*5 also the derivative is 2b -4 = 0 so b = 2 is the minimum since second derivative is positive but that doesn't help either. What do I do?

You can't just arbitrarily remove numbers (3 in your case) in the factored part of the equation and plugging them into another equation. Just think that if you remove a number from factored part of the equation, you are affecting the whole factored part of the equation with that number, in other words that number has an effect on all parts of the factored part of the equation.

Let f(x - 2) = x^2 + 2 Find f(3) Do I just sub 3 for x and get f(1) = 11? or do I do it this way: x - 2 = 3 x = 5 , therefore f(5) = (5)^2 + 2 = 27? or just let x = 5, then f(5 - 2) = f(3) = 5^2 + 2 = 27?

Personally I find the above line hilarious especially given the fact that he asked this question several times on multiple forums.

Thanks Doc. P.S. I guess I felt that I cheated because the problem seemed too simple, also I don't know how to include the free body diagram so everybody can see it that's why I tried to explain the best I could. I'd appreciate if someone could tell me how I could do that.

OK, thanks imatfaal.

Right, v, v0 and x is all we have given, didn't really think this one through, sorry.

First, since you know velocity solve v = v0 + at for time. Then, use x = x0 + v0t + 1/2 at^2 solve this for acceleration and the figure out what it is considering the gravity is 9.8 m/s^2.

I have 3.43 GPA at the community college, because in the beginning I didn't care about my GPA and got a C in Biology and an F in Chem I and a couple of B's, but I retook Chem I and got all A's since then. Will my low GPA affect my chances to get to medical school? I am planning to get all A's and at the 4-year University, but I'm not sure if they will count my GPA form community college.

Don't worry about it I solved it already.

Ok, I did it. Thanks a lot for your help I really appreciate it.

Great, thank you very much DrRocket, I gotta run now but I'll try to solve this a little later. This problem is given in the first semester Physics course and unfortunately before we learned about rotational motion besides the ar = v^2/r , I'm undergrad.