| Problem statement Solution video |
DISCUSSION THREAD
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DISCUSSION
Shown in the animation below is the path taken by Particle P.
- As expected the unit vectors et and en are tangent and normal to the path of P, repectively.
- Also seen there is that the velocity vector is always aligned with the tangent unit vector, et, since v = v et.
- The acceleration vector, in general, has both tangential and normal components of acceleration. The normal component always points in the same direction as the unit normal vector, en. The acceleration vector either points "forward" of the motion (for which the rate of change of speed is positive; i.e., increasing is speed), or points "backward" of the motion (for which the rate of change of speed is negative; i.e., slowing down).
HINTS
Recall that since you do not know the y-component of motion of P explicitly in terms of time, you will need to use the chain rule to find dy/dt.
It is most convenient to determine the rate of change of speed through the vector projection of the acceleration vector, a, onto the unit tangent vector, et. How do you find the tangent unit vector? Simply divide the velocity vector by its magnitude! et = v/v .
Once you know the rate of change of speed, you can find the radius of curvature ρ through the magnitude of the acceleration.
Did anyone else find that acceleration didn't have a tangential component and velocity didn't have an er vector? Since both these unit vectors must be perpendicular it would make sense but, I am concerned that this may be due to improper differentiation.
Please disregard - this was meant as a comment for 1.E
Are the velocity and acceleration for part b supposed to be in cartesian or path description?
Beings that all of your given information is in Cartesian, providing the answers in Cartesian would make sense.
part b is cartesian.
part d is path description,
btw: It is most convenient to determine the rate of change of speed through the vector projection of the acceleration vector, a, onto the unit tangent vector, et. How do you find the tangent unit vector? Simply divide the velocity vector by its magnitude! et = v/v .
Once you know the rate of change of speed, you can find the radius of curvature ρ through the magnitude of the acceleration.
We are only asked to find the rate of change of velocity (v dot) in part D, I believe that we do not need to find the radius of curvature.
My sketch of the path of P has the vertex of the quadratic function in the third quadrant of the XY-plane. However, the animation/visualization represents the vertex in the 4th quadrant. Is the animation just for general reference and not exactly representative of the path we are given?
If you graph the given equation in Desmos, you'll see that the vertex should be in the 4th quadrant.
When they ask for rate of change of speed how is that different than finding the acceleration that we find in part b?
I think that means they want you to find the tangential acceleration.
I think the problem statement saying "find rate of change of speed" is just trying to tell you to find path components for acceleration.
It is most convenient to determine the rate of change of speed through the vector projection of the acceleration vector, a, onto the unit tangent vector, et. How do you find the tangent unit vector? Simply divide the velocity vector by its magnitude! et = v/v .
Once you know the rate of change of speed, you can find the radius of curvature ρ through the magnitude of the acceleration.
Is there any specific reason why the value for x is given?
I believe since P's velocity and acceleration have a y-component as well, they gave us x and x_dot as our parameters to solve for y, y_dot, and y_dot_dot numerically.
I think I might be on the right track with this, but I'm not sure. For part B, do we need to use the equation for the radius of curvature at x = 12 in order to find the acceleration? But then using the equation a = v_dot * e_t + (v^2/p) * e_n I am lost as to how to find v_dot. With the way I solved for v it doesn't seem possible.
I don't think you need to use the equation for radius of curvature. You can solve it like that, but I think doing it in cartesian would be easier.
The rate of change of speed is v_dot right? Then v_dot = e_t * a right?
I think I am getting thrown off by units. For y_dot, I have (feet - no unit) * ft/s but I know it is suppose to be ft/s. Can I just drop the feet unit?
For y_dot_dot, I have 0 + ft/s when it should be ft/s^2. I am not sure if I can drop units and just do the math then add them back or not.
I think it's the derivatives messing me up. I keep wanting to use the derivative of x as 1. Will the derivative of x always be x_dot in this class? Even still, my units are messed up.
x_dot is a constant. This is the derivation of x with respect to t
I am also finding that my units for y_dot are ft*ft/s. I know this isn't right but I'm not sure what I am doing wrong.
The numeric coefficients in the expression for y(x) have units embedded in them. In the end, x and y both have units of feet.
When doing part D for this problem. Is anyone ending up with clean round numbers or are you getting answers with decimal places? I am not sure if I messed up somewhere.
For part D I got an answer with decimals. I'm pretty sure you use the equation a*et = v(dot).
For part D, once I have the equation for the Rate of Change of speed of P, do I need to plug in the same point x=12? I would assume so but I just want to make sure
For part D I am using the equation for v_dot from Example 1.C.2. Is that the correct application of this example?