How Do I Solve These AP Physics Problems?

[tdub96]

Hey guys, I am new to the forums. I am currently taking AP Physics in my senior year of high school, and I am having some difficulty solving the following two problems:

1. A ball is shot upward with a velocity of 30ft/s. Another is shot upward a second later at 100ft/s. Where and when do they pass?

2. The equation of motion of a specific particle is x=6t^2+4t^3. X is in feet, t in seconds. Compute the objects initial acceleration and its acceleration when position is 100 feet.

My teacher is infamous for not teaching us how to do most of the stuff, but I was able to figure out the homework minus these two questions. If anyone can help that would be extremely appreciated!
Thanks in advance...
-TW

 

[QuarkCharmer]

For #1 they are basically asking you to find the intersection between two parabolas. You can use the kinematics equations for each ball and then set them equal.

#2
The initial condition is when the time is 0 right? They tell you the other condition is when the time is 100. What you need to do is figure out how to get a function for acceleration and/or velocity out of the position function they provided you. Isn't there something you can do to the function that outputs another function giving the slope of the first at any t?

 

[tdub96]

What kind of kinematics equations for #1?

And yes, for #2 the initial time =0. So then the initial acceleration would be 12ft/s^2.

By using derivatives, the equation for velocity would be 12t-12t^2, and acceleration would be 12+24t. I just don't know where to go from there.

 

[QuarkCharmer]

Okay, so you found the second derivative of the function to be this right?

a = 12 + 24t

From that you put in t = 0 to get your initial acceleration of 12. Now they want to know what the a is when the position is 100 right? You need to find the time when position is 100 and put that into the second derivative. You can find that from the original function right?

For the first problem, simply find the kinematics equation that relates initial velocity to position. Remember that one is being shot up a second later. You know about the transformations of basic graphs right?

 

[tdub96]

Correct, so I'd set 100=6t²+4t³. By simplifying, I'd get t³+2/3t²-25=0. How would I factor that to solve for t?

As for kinematics, there are a couple velocity and position related functions. X=1/2aΔt²+V_o+x_o and v=√(v_o²+2aΔx). I am assuming to use the first one?

 

[QuarkCharmer]

Well, for AP physics I think they might want you to simply use a calculator to graph it and find t when x = 100. If not, then you can use the P/Q test to find a zero, then use polynomial long division to start breaking the function down.

\Delta y = V_{iy}t + \frac{1}{2}A_{y}t^{2} is a pretty nice kinematic equation isn't it? (It's almost like your first one, but I think you are missing a t in there somewhere)

 

[tdub96]

X=1/2aΔt²+V_ot+x_o, yes I was missing a T. Would it be logical to set two of these equations equal to each other like this...

1/2aΔt²+V_ot+x_o=1/2a(Δt+1)²+V_o(Δt+1)+x_o

Or am i better off using your equation? I've never seen the one you posted before...

 

[QuarkCharmer]

The one I posted is basically the one you posted.

\Delta y = V_{iy}t + \frac{1}{2}A_{y}t^{2}

y_{f} - y_{i} = V_{iy}t + \frac{1}{2}A_{y}t^{2}

y_{f} = V_{iy}t + \frac{1}{2}A_{y}t^{2} + y_{i}

Just different symbols. You have X for my y_f and I have y_i for your X_0.

If you want to shift the graph of the ball that started late to the right, should you be adding to the dependent variable or subtracting from?

 

[tdub96]

I should be adding one to the T of the second graph as it is launched a second later, correct? Then, I would graph each equation and the intersection of the two parabolas is the time when the two objects pass mid-flight?

 

[tdub96]

Anyone?