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(II) Show that $\overline{v}=\left(v+v_{0}\right) / 2$ (see Eq. 12 $\mathrm{d} )$ is not valid when the acceleration $a=A+B t,$ where $A$ and $B$ are constants.

$\bar{v} \neq \frac{1}{2}\left(v+v_{0}\right)$

Physics 101 Mechanics

Chapter 2

Describing Motion: Kinematics in One Dimension

Physics Basics

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Cornell University

Simon Fraser University

University of Winnipeg

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This is quite an interesting problem. It's very theoretical. So we want to show that this equation for the average velocity where's some of being V zero over to? We want to show that this is not valid when we have an acceleration of this form A plus b t where a and B are constants. So this this acceleration is represents exploration. That's not constant. And to show that the this average velocity equation doesn't work here, we're gonna need a little bit calculus. So I know that I can say aye is d X, not X TV, B t. And then we can say from that that are Devi is a d t pretty simple. And now we need to integrate. So I'm gonna integrate Devi and A T. T and the bounds of this inter girl. I want to go from V zero our initial velocity to V and I'm going to go from a time of 02 time t And from here we just get V going from V zero to V and on the right side, we want the integral from zero to t in. Our acceleration is a plus. B t DT. So the left side just comes out to V minus Be zero on the right side comes out to a T plus 1/2 B t squared And now I can find an expression for the velocity Gonna be v zero plus 80 plus 1/2 feet Peachy is great And now do the same thing that I just did, but with velocity. So I know that velocity is dx DT. And so I can say that d x is VD p andl integrate once again and the bounce of my integral I'm going to go from a position x zero to x and again from time zero to time t So just like the last in the grill, we're going to get X minus x zero and our velocity function. It's what we solved The last part be zero plus 80 plus 1/2 beachy squared. So this integral comes out to be, uh, let's see, B zero The first term we had a T second term is gonna be 1/2 a T squared this term t squared Integral is gonna be 1/3 t cubed and 1/3 time's 1/2 is 1/6 b t cute and so I can get an expression for a position is x zero plus B zero t that this should be a plus plus 1/2 a T squared plus 1/6 beachy cute. And now I'm gonna go back to this. What we're trying to show, uh doesn't work. So I have V equals our average V is V plus B zero over, too. And to make things easier, so we have this function for position and this function for velocity. I'm just going to set, uh, V zero and zero 20 Because why not? It doesn't make a difference. And so then we're gonna get for our average velocity. I'm going to get V over, too. Just gonna be a t plus 1/2 p d squared over two. We get 1/2 80 plus one over four b t squared. So that's what we get if we use this equation for average velocity and then we want to show that that doesn't work. So we know that average velocity is always going to be Delta X over Delta T. This is always true. And so if we plug in Delta X here Ah, actually, we don't even need X zero to be zero because we have Delta X. It doesn't matter, but we still know v 00 So this is gonna be 1/2 80 squared plus 1/6 b t cubed over our delta Tea is just tea because you're going from zero to T. And so this is 1/2 a T plus 1/6 B t squared. And now you can see we have this value for first equation just not equal to what we get when we used all the ex of adult teeth. So that says that when our acceleration isn't of the form A plus B t which is not constant than this equation here, V equals half of V plus B zero does not work.

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