Classical Mechanics and Net Displacement

[Nicol]

Hello, people of the physics forums. I'm trying to get a grasp on physics, and I have come to my first real roadblock. I'm looking at Classical Mechanics, and more specifically, net displacement.

On http://farside.ph.utexas.edu/teaching/301/lectures/node20.html, I have attempted a set of questions, which are answered. I have completed the first two, and do not understand the final one. I don't know what a v-t curve is, and how to find net displacement of this, although I have been searching online for a good week now. I have attempted to get the areas for the various shapes within the graph, but to no avail.

Any help would be appreciated.

 

[Nicol]

You know, I really should have put this in the "Classical Physics" section. My mistake.

 

[cristo]

Is it just me, or is the answer not given on that page? A v-t, or velocity-time graph is a graph of velocity plotted agsinst time, like that one at the top of the page. The first equation in the solution says that v=dx/dt, and so we see that the net displacement, x, is

[tex] x = \int_0^{16}v(t)dt [/tex] shown in the second equation. The integral of a function, is simply the area under the curve, so calculating the area under the curve of the v-t graph at the top of the page gives the result.

I'm not sure whether this helps you, as I've more or less just stated what's in the solution!

 

[Nicol]

Yeah. I don't actually understand the formula. I don't understand what the variables in the formula represent, and

 

[Nicol]

Yeah. I don't actually understand the formula. I don't understand what the variables in the formula represent. I don't understand exactly where the curve of the v-t graph is.

This is what I get for being interested in physics after I have left school.

 

[cristo]

Ok, well the curve of the v-t graph is just the line plotted (mathematicians often call straight lines "curves") In the formula, x is the displacement, v is the velocity and t is time. The formula v(t)=dx/dt just says that velocity is the rate of change of position. Have you come across calculus before? If so, integrating this expression yields the formula [tex] x = \int_0^{16}v(t)dt [/tex] where the numbers on the integral sign are the intial and final values of t.

 

[Kurdt]

I think at that level it is very misleading to introduce calculus. For one thing the curve is not continuous which adds minor complications. Secondly it is easier to state that the distance traveled is really the area under the graph and I'm sure you can work out areas of basic figures such as squares and triangles. I'd say just ignore the calculus part until you have a deeper understanding of the mathematics involved.

So in summary the distance traveled on a velocity - time graph is mearely the area under the curve.

 

[Nicol]

Thankyou for your help.