Answer

A web page designer creates an animation in which a dot on a computer screen has a position of $\vec{r}=<4.0 \mathrm{cm}+$ $\left(2.5 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2} > \hat{\boldsymbol{\imath}}+(5.0 \mathrm{cm} / \mathrm{s}) t \hat{\boldsymbol{J}}$ (a) Find the magnitude and direction of the dot's average velocity between $t=0$ and $t=2.0 \mathrm{s} .$ (b) Find the magnitude and direction of the instantaneous velocity at $t=0, t=1.0 \mathrm{s},$ and $t=2.0 \mathrm{s} .$ (c) Sketch the dot's trajectory from $t=0$ to $t=2.0 \mathrm{s},$ and show the velocitiescalculated in part (b).




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Video Transcript

So what we have here is a dot on a screen given by this position vector right here and for a what we need to do is find the average velocity. Uh, between time equals zero and time equals two seconds. So the average velocity is going to be given by the final position, minus the initial position over time. So, uh, for a time, we No, that is from 0 to 2. So what time's just gonna be to sex? For the final and initial, we're gonna have toe look values in. So the final is at time equals zero. I mean, time you calls to just equal to four centimeters plus t squared, which is two squared. So 44 times 2.5 is 10 in the I had direction, plus five centimeters per second times. Time of two is going to give us 10 centimetres in the J have direction. Now our position vector at T equals zero. That's going to give us four centimeters in the I had direction plus zero in the J had direction. So now you just subtract them and you get 10 centimeters in the I had direction less 10 centimeters the Jay had direction all over too. And you can divide to from both those and you're left with five centimeters in the I had direction in five centimeters in the J a happ direction. So now we find the velocity seeing triangle. So this the I had direction. It's the J had direction five and five by the magnitude we used staggering here. Um, so the magnitude or the average velocity it's going to equal five squared plus five squared elsewhere rooted, and that is going to give us an average velocity of 7.1 centimetres her second. Now, to find the magnitude, we just use our trick function. We're, I mean, not magnitude. Uh, angle. We're looking forward to single right here. So we're going to choose Tan data. He cools 5/5 inverse tangent of one is data equals 40 five degrees part B. We need the instantaneous velocities at time equals zero. Time equals one, and time equals to now to find the instantaneous velocities. Uh, we need to use the velocity vector, and the velocity vector is going to be given by the time derivative of our position back. So just taking the derivative. We have the velocity vector equal to two times 2.55 centimetres per second Squared times, time in the eye hat direction plus jay had direction. We just have five centimetres per second. So now we just need to find the velocity. Time equals it's your So we so t in the eye direction That's gonna give us zero plus just five centimetres per second in the J had direction, remember? So there's zero in the eye direction. So that means, uh, essentially and we have this but this vector right here her this line is going to be zero. That means this angle right here is just gonna be 90. So Alfa equals 90 degrees and our magnitude of the velocity is 25 senate meters per second for velocity at time equals one. We have five centimeters. Her second in the I had direction plus five centimeters per second in the J hat direction. Now this is similar, Teoh what we have appear It's the same numbers, right? But it's not the same problem, but for self that So you know that the direction here is going to be 45 degrees and our velocity it's going to equal 7.1 centimetres per second now for the last one. Velocity of time equals two. We're going to have 10 centimetres per second. Plus in the I had direction, of course. Five centimeters. Her second in the J had direction. Now we have to set up our triangle in the eye direction I had direction is 10 and in the J have direction. We have five looking for this diagonal. Can we just use private Pythagorean theorem? So the V two equal five squared plus 10 squared all that square rooted and we end up with a velocity of 11.2 centimetres per second. Now I find that data we can use tangent tangent or we'll call it Alfa Tinge in Alfa equals opposite over adjacent. So we have that Alfa here is equal to 26 0.6 degrees. Now the last thing we need to do is plot the trajectory on X Y plane from time zero to time equals two. And so what we'll need is the position at time equals zero. Time equals one and time equals to help us. So at the position, vector of time equals zero. We have four centimeters in the I had direction plus zero in the J hat direction. That means we are starting at 123 four here for there are two me equals one. We have four plus 2.5. So that's going to be 6.5 centimeters. And I have direction plus five centimeters in the JM direction. That's gonna be 12345 6.5. It will go up to about five se right around there now or hat Well, T equals two. So we have four plus 2.5 times four. So we have 14 centimeters and I had direction. Plus, she does find its 10 centimetres in the J hat directions 6.5789 10 11 14 se Right around here, where it ends up, so are lost ease. We have that we have. So you know it goes. It starts at a 90 degree trajectory turns and at this point is that 45 and then here it's at 26.6. So here we have the velocity going upwards. That five centimetres per second here is 45 degree angle at 7.1 centimetres per second and here it angle.

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Look out 23 26.6. We have a velocity 11.2 centimetres per second.