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Physics Problems.
Basic
Registered User regular
Registered User regular
1) A car traveling at 23.0 m/s runs out of gas while traveling up a 19 degree slope. How far up the hill will it coast before starting to roll back down?
For number 1, I understand that I need to plug in the values for the equation 1/2m(v^2)=mgh, and since mass (m) is on both sides I can cancel them out. However, no matter how many times I do this calculation, I get something over 180 m before it rolls back down, when I am told it should be more like 82.9 m. This is what I do, for clarification:
1/2(23m/s) = 9.8m/s^2(h)
529/2 = 9.8m/s^2(h)
264.5/9.8 = h
26.9897 = h
h = d(sin 19) where d is the distance traveled by the car along the incline
26.9897/(sin 19) = 180.07 = d
I don't understand why I am getting a different number, can anyone explain what I'm doing wrong?
2) The figure shows the velocity graph of a train that starts from the origin at t = 0s.
Draw position graph for the train.
I know that velocity is an derivative of position, so position must be an integral of velocity. What I don't understand is why after plotting the integral at t_final = 1s and t_initial = 0s I get funky answers (I am supposed to get a curve that is the negative y-axis from t=0 to t=8).
Can anyone show me what I am doing wrong? Perhaps I am not integrating incorrectly?
Also, I need to know the steps. I am trying to understand the material, and I can't do it if I'm only given the answers again.
Any help would be appreciated.
For number 1, I understand that I need to plug in the values for the equation 1/2m(v^2)=mgh, and since mass (m) is on both sides I can cancel them out. However, no matter how many times I do this calculation, I get something over 180 m before it rolls back down, when I am told it should be more like 82.9 m. This is what I do, for clarification:
1/2(23m/s) = 9.8m/s^2(h)
529/2 = 9.8m/s^2(h)
264.5/9.8 = h
26.9897 = h
h = d(sin 19) where d is the distance traveled by the car along the incline
26.9897/(sin 19) = 180.07 = d
I don't understand why I am getting a different number, can anyone explain what I'm doing wrong?
2) The figure shows the velocity graph of a train that starts from the origin at t = 0s.
Draw position graph for the train.
I know that velocity is an derivative of position, so position must be an integral of velocity. What I don't understand is why after plotting the integral at t_final = 1s and t_initial = 0s I get funky answers (I am supposed to get a curve that is the negative y-axis from t=0 to t=8).
Can anyone show me what I am doing wrong? Perhaps I am not integrating incorrectly?
Also, I need to know the steps. I am trying to understand the material, and I can't do it if I'm only given the answers again.
Any help would be appreciated.
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It does help if you integrate over the appropriate domain.
As for the first one, looks to me like you're taking the sin of 19 radians instead of degrees.
EDIT: So you don't think I'm sneering at you or anything, I spent many hours of the senior year of my aerospace engineering degree both making these mistakes and correcting them for other people in a bevy of situations. If you ever intend to use physics, it is terribly terribly important that you pay attention to these details, or they will be the end of you and possibly the people who use your bridge.
Set your calculator to the correct setting for degrees. It is probably on radials.
sin 19* = 0.326
sin 19rad = 0.150
And the second question, the train starts of with a negative speed. This means it's moving towards the left (if x is on a left to right track). It's slowing down, and at t = 4s it's standing still, up until t = 8s, when it hits top speed in the other direction, after which it plods on. So yes, if you start the train at x,t = 0,0 it will move negative for 8 seconds, after which it returns to it's starting point, and goes forward from that point.
I don't understand what you mean when you bolded over the domains. Are you saying that using t=0 and t=8 as domains is what I am supposed to be doing? Because I don't understand how that will help me plot a graph. I am not using a calculator to do that problem by the way, since it's hard to tell where the points are on one and I need to actually draw the graph.
v(t) = 1/2 t - 2 for 0 <= t <= 8
v (t) = 2 for 8 < t <= 10
Integrating this would yield
x(t) = 1/4 t^2 - 2t + c1 for 0 <= t <= 8
x(t) = 2t + c2 for 8 < t <= 10
With c1, c2 as constants, where it's obvious that the two should neatly cross over (So c2 = c1-16).
This is a parabole from 0 to 8, then a a straight line with 2 slope from 8 to 10.
ps: I'm old school and predate graphic calculators. This is how I was taught to do it.
EDIT: Fixed a minus sign mistake. No more math after midnight!
You said you used t_final = 1 second. What inspired you to do that? The only times that matter (i.e. things change) are 0 seconds (things start), 8 seconds (things change), and then the end (10 seconds). Well, there's another time that matters, but I'll leave it up to you to figure out what time that is.
You have to integrate this function--which isn't discontinuous anywhere, by the way, it just isn't differentiable at t=8--in pieces. From 0 to 8 seconds, speed is steadily increasing from -2 m/s, and you'll have a nice little sinking parabola for the train's displacement. Beyond 8 seconds, the speed remains constant, so the displacement will make a nice little line.
I can practically guarantee that the things you're supposed to recognize in number 2 are:
- I have to integrate velocity to find displacement
- Integrating a sloped line gets me a parabola, and I need to know where the extreme (minimum in this case) of that parabola is
- Integrating a constant gets me a sloped line
You plot the points that matter on the graph and then sketch basically what things should look like.
Sounds like you need to brush up on your integration.
I understand how integrals work but I never knew that integrating a slope means I get a parabola, and that integrating a constant gets me a sloped line.
Think backwards from derivatives.
You get the derivative of a slope you get a line
you get the derivative of a 2nd order quadratic, you get a sloped line
ect, ect, ect.
To give you an idea of how important calculus is to Physics: it's why calculus was invented. At least, that's why Newton did it; I'm not as familiar with Leibniz' story.
I'd recommend getting comfortable with simple integration and differentiation quickly so you can concentrate on the physics concepts in physics. For the problem you showed us, you can get a dandy sketch of the displacement by just knowing the following:
- displacement is the integral of velocity (which you seemed to recognize)
- the integral of a curve is equal to the area between the curve and the zero line
- the velocity function of the train forms two triangles and a rectangle w.r.t. zero velocity
- the area of a triangle is one half the area of a rectangle of equal length and height
- the area of a rectangle is its length multiplied by its height
You don't even have to actually find the equations of the lines. Just looking at the graph, I can tell you the maximum negative displacement, when it occurs, and what the displacement is at every tick along the time axis of the graph.
Look ahead in your calc book and see if it doesn't have a section explicitly about sketching curves from their derivatives/integrals. All of mine did, and this is something I was required to do in some form or other in basically every engineering course I ever took.
that's why we call it the struggle, you're supposed to sweat
To get position from velocity? You're integrating somehow, even if it's not symbolically or even formally.