# Equation of Kinematics for Constant Acceleration

Well, my first post here and it happens to be in the homework section. =P I finally escaped AP Chemistry (and got a 4 on my AP test!), but because I chose to take the AP Physics course for the upcoming school year, I undoubtedly have summer work. The task seemed simple enough: the teacher handed us a sheet of equations, and we locate them in the book; then we write down some information and provided an example problem.

Unfortunately, I have run into some obstacles.

On the equation, x = x0 + v0t + 1/2at^2, an equation of kinematics for constant acceleration, I am unsure about the answer to my example problem. It'd be wonderful if someone could flat-out tell me if my answer is correct or not, and it'd be even more wonderful if someone could point out where my mistake was.

Problem: A car is traveling at a constant speed of 27 m/s on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain, so that the two cars meet for the first time at the next exit, which is 1.8 km away?

Work: 27 m/s x 1/1800 m = 1/.015 s = 66.7 seconds.
First, I divided (sorry!) the given speed and the distance to obtain the seconds.

x = x0 + v0t + 1/2at^2
(1800m)= (0) + (0) + 1/2a(66.7 s)^2
Then I plugged in the corresponding numbers.

a = .81 m/s^2
Finally, I solved for a.

I hope I did it right... I'm hesitant on whether I plugged the right things in. And I'm sort of hesitant on whether the equation is right for the problem. =\

Moving on... I located 78/80 of the equations needed (yay!), but the last two, I cannot seem to find anywhere. I googled it, I asked other people, but to no avail. So I'm wondering if anyone could identify it and simply provide me with the name of the equation.

One of the equations is U = -Gm1m2/r.
U being the potential energy, G being the universal gravitational constant, m1 and m2 being masses and r being radius or distance. I can't find it anywhere.

However, I found an equation similar to it: F = -Gm1m2/r^2. What is the difference between these two? Is there any difference at all?

The final equation I can't locate is: V = 1/(4 pi ε) Σ qi/ri.
ε, according to the key is emf, q is point charge and r is distance. I thought that it would be a series equation, but alas, it was nowhere to be found in that chapter.

I apologize for this post being so lengthy, but I hope that someone might be able to aid me in my work. [=

Last edited:

Doc Al
Mentor
User Name said:
Work: 27 m/s x 1/1800 m = 1/.015 s = 66.7 seconds.
First, I multiplied the given speed and the distance to obtain the seconds.

x = x0 + v0t + 1/2at^2
(1800m)= (0) + (0) + 1/2a(66.7 s)^2
Then I plugged in the corresponding numbers.

a = .81 m/s^2
Finally, I solved for a.
Looks OK to me. But when you calculated the time, what you did was divide the distance by the speed, not multiply. That comes from:
x = vt, of course.

One of the equations is U = -Gm1m2/r.
U being the potential energy, G being the universal gravitational constant, m1 and m2 being masses and r being radius or distance. I can't find it anywhere.
That equation tells you the gravitational potential energy between two masses, as you know.

However, I found an equation similar to it: F = -Gm1m2/r^2. What is the difference between these two? Is there any difference at all?
That describes the gravitational force between two masses. Very different! It's called Newton's law of universal gravity.

The final equation I can't locate is: V = 1/(4 pi ε) Σ qi/ri.
ε, according to the key is emf, q is point charge and r is distance. I thought that it would be a series equation, but alas, it was nowhere to be found in that chapter.
That's the electric potential due to a collection of charges, where ri is the distance from charge qi to the point in question. (ε is the permittivity constant.)

Ah, thank you very much for clarifying a few things.

The electric potential equation.. I think I may have found; someone told me this was it, but I wanted to double-check: V = EPE/q0.
V being the electric potential, EPE being electric potential energy and q0 being a test charge.

Is it the same equation?

The equation that tells the gravitational potential energy is nowhere in the book. I can't seem to find it. When I looked up gravitational potential energy, I got the equations:

W = mgh0 - mghf

and

PE = mgh

But I doubt any of those are equal to the gravitational potential energy equation.

Doc Al
Mentor
User Name said:
The electric potential equation.. I think I may have found; someone told me this was it, but I wanted to double-check: V = EPE/q0.
V being the electric potential, EPE being electric potential energy and q0 being a test charge.

Is it the same equation?
Those are equivalent ways of expressing the same thing. For more on the relationship between electric potential vs potential energy, see this: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elepe.html

The equation that tells the gravitational potential energy is nowhere in the book. I can't seem to find it. When I looked up gravitational potential energy, I got the equations:

W = mgh0 - mghf

and

PE = mgh
These equations are used to find changes in gravitational PE (or work done by gravity) near the earth's surface. (Since the earth's gravity gets weaker with distance, those equations can't be used for distances too far from earth.) Note that the zero point for gravitational PE (where h = 0) is arbitrary, since only changes in PE make a difference.

That other equation for gravitational PE is a much more general expression for the PE between any two masses. Note that that equation takes the zero point to be at infinity. For more about gravitational PE, again I will refer you to hyperphysics: http://hyperphysics.phy-astr.gsu.edu/HBASE/gpot.html