Recent content by Wecht

So if it's an x vs t graph of the car, it doesn't matter where the person is. Why is TinyTim suggesting that the car is falling out of the sky? And also that the origin is the position of the person?

Wait, is this supposed to be an x-position vs time graph of the car, or what?

I don't know what you're saying; are you pointing out a way to solve it using PE? Is the way I used wrong?

I think you guys are making this more complicated than it really is.

Here's how I would do it from 1000m to 200m upwards force=50N downwards force=W=mg=784.8N so net force is 734.8N downwards. F=ma solve to get a (from 1000m to 200m), which I found to be -9.185m/s^2 Vfy^2 - Viy^2 = 2a(yf-yi) solve for Vfy, for which I got 121.2m/s (at 200m) However, if...

I'm not sure I'm understanding you correctly. The question says the car is coming down the street towards you; I'm pretty sure it's safe to assume that it's not falling out of the sky.

The second way you put it is correct; as with the velocity, you must translate the acceleration into terms that fit your coordinate system, as -y-direction is no longer the only direction of acceleration.

Remember that the equation f = (mV^2)/r is a condition that must be satisfied for centrifugal force; it is not a law.

I'm also assuming that by 'x-coordinate is now 10cos30degrees' you mean x-velocity, and the same thing for 'y-coordinate'. Also realize that 'g' is no longer -9.81m/s^2 y-direction, if you've tilted the axes. Another point, don't you need to know what height above the incline the projectile...

BUT, you have to realize that the info bit 'horizontal' in the 'horizontal velocity' they give you is independent of your chosen coordinate system; meaning that if you 'tilt the axis', 'horizontal velocity' is no longer synonymous with x-velocity. What I mean is, you can change the coordinate...

Yes, absolutely you can. Designation of the axes and is arbitrary, meaning that you can choose the origin and the direction the axes point. Though it's probably a good idea to make sure they're perpendicular to each other.

Hm I don't think so, TinyTim. The origin should be the car's position at t = 0. Since it's accelerating towards him, he can't be at the point where the car is, because then it wouldn't be accelerating towards him.

How did you shape it? If it's curved like the letter 'J', I don't see what's wrong with that. But if it's shaped like the letter 'r', that means it decelerating. Do you know for sure that the curve is wrong, or did you maybe label the axes incorrectly, or not put a title for the graph?

I'm also taking Physics 111, and I have some similar problems. What I do is refer back to my notes from Physics 30 and Physics 20 (Grade 11 and 12 physics). The simpler concepts might have been better explained to you back then compared to any explanations you might get from university...

Slope of zero on a velocity-time graph shows that the velocity is constant at that point in time. Slopes on velocity-time graphs are representations of acceleration. positive slope = positive acceleration = negative slope = negative acceleration slope of zero = zero acceleration =...