A web page designer creates an animation in which a dot on a computer screen has position r→=[4.0 cm+(2.5 cm/s2)t2]iî+(5.0 cm/s)t jĵ. (a) Find the magnitude and direction of the dot's average velocity between t = 0 and t = 2.0 s.

Video transcript

Welcome back everybody. We have a screensaver on a computer screen whose position is given by this function of time. And we need to find the average velocity, both its magnitude and direction between this time interval, let's define a few key things here first, our average velocity is going to be given us given to us by our change in our over our change in T. But what's our change in our in this case we are changing our is going to be our final our minus our initial R. And r. Change of T is going to be similar where it's gonna be our final I'm minus our initial time. So let's go ahead and find those values first. Well, our final are is going to be well we're just gonna plug in our final time into our equation here. So let's go ahead and do that. We're gonna have five plus 1. times three squared three being our final time. And this is our X component. So we're gonna multiply this by the unit factor I Plus four times final time three And this is going to be our y component. So multiply that by the unit factor J. This is equal to 12. I plus 12. Okay, great. Let's go ahead and find our initial R. The same thing here. Except we're just plugging in this value of T instead our first time. So we're gonna have five plus 1.8 times one squared. This is going to be our X component Plus four times 1. And this is going to be our J component. Now, this is going to be equal to 6.8 I plus four K. Now let's go ahead and find delta R which once again is just R f minus R I. And we're gonna do this component wise. So for our X component we will have 12.2 minus 6.8 Plus. For y component we will have -4. This is going to give us 5.4 I plus eight day. So now that we have our delta R, let's use that in combination with delta T to find our average velocity. So we have our average velocity is equal to delta are over delta T. So we have 5.4 I plus A J divided by delta T. While the change in time is just going to be three minus one which is two. So this is equal to 2.7 I Plus 4.0J. Great. So now that we have our average velocity we are going to need to find the magnitude and direction. So let me give some helpful formulas for this magnitude of a given vector. A given vector A is going to be the square root of its X component squared plus its y component squared in the direction that we are looking for represented by the angle theta is going to be equal to the arc tangent of its Y component divided by its ex opponent. Well we have the X component right here and the Y component right here. So let's go ahead and find those values first and foremost we're going to find the magnitude of our average velocity. This is going to be equal to the square root of its X component. 2.7. Where plus its Y component Four squared when you plug this into your calculator you get 4.83 cm/s. Now let's go ahead and find that direction that we're looking for. So theta is equal to the arc tangent of its Y component or divided by its exponents of 2.7. When you plug this into the calculator, you get that. This is 56 degrees counterclockwise above the X axis. So now we have found both the magnitude and the direction of our average velocity corresponding to answer choice C. Thank you guys so much for watching. Hope this video helped. We will see you all in the next one.