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A puck is moving on an air hockey table. Relative to an x, y coordinate system at time t 0 s, the x components of the puck’s initial velocity and acceleration are $v_{0 x}$ 1.0 m/s and ax 2.0${m} / {s}^{2}$ . The y components of the puck’s initial velocity and acceleration are $v_{0 y}$ 2.0 m/s and $\mathcal{U}_{y}$ 2.0${m} / {s}^{2}$ . Find the magnitude and direction of the puck’s velocity at a time of t 0.50 s. Specify the direction relative to the x axis.

$27^{\circ}$

Physics 101 Mechanics

Chapter 3

Kinematics in Two Dimensions

Motion in 2d or 3d

University of Michigan - Ann Arbor

University of Washington

Hope College

Lectures

04:01

2D kinematics is the study…

10:12

A vector is a mathematical…

03:38

Two identical pucks are on…

03:06

A rocket-powered hockey pu…

04:38

A 0.160-kg hockey puck is …

04:49

01:19

A hockey puck is traveling…

06:14

A hockey puck with mass 0.…

14:28

05:25

A0.160-kg hockey puck is m…

05:18

A 0.25 kg puck is initiall…

03:01

A hockey puck of mass $150…

18:37

Two identical pucks colli…

02:00

05:50

Alice tapes a small, $200 …

04:02

Position and velocity from…

12:13

The position of a moving h…

13:54

Hockey puck $B$ rests on a…

00:34

In a slap shot, a hockey p…

00:59

02:10

A particle has a constant …

Okay. Welcome. We're gonna do problem. Where? 16. Here. Oh, so first off, it gives us two initial conditions, possess an initial speed, and the ex were gonna write V initial on the X is one meter per second meter per second, and it gives us an initial velocity and the why of two meters per second. It also gives us an acceleration. The axe. So the acceleration The X, it glows. Uh, two minutes per second squared and acceleration. The while was negative. Two meters per second squared. And this is over some time period. Delta T equals 0.5 seconds. That's asking, uh, what is the asking? What is the final angle on the X axis? After all this is finished. So we want to find initial velocities for our X and y so we can do that. Basin, the formula. Initial finals lost in the axes, the initial lost in the ax plus acceleration. The X times the change in time because we know lost equals acceleration, times time. So here, we're gonna get, uh, one meter per second, plus the acceleration of 2.0 meters per second squared times the change in time, which is 0.50 seconds, which is gonna leave us with one plus one which overall is two meters per second in the ex direction. And we knew something for Why so v f why is equal toe the initial velocity and the Y plus acceleration? Why times of change in time. We know that's two meters per second. Plus this is negative. Noticed the acceleration. Why is negative two meters per second squared times are changing time of 0.5 seconds. So this is really like two minutes per second minus one because negative to half our sorry negative two meters per second divided by two is the same as multiplying by 0.5 is negative one So two plus native one is our final velocity is one meters per second and why? Okay. Perfect. So then we have This is our axe we have this is there. Why? So what's our final angle going to be? You know, starting a new page, a new white word. So, uh, we used two is our axe. One is or why we want to find this angle here. Sita. So he said this is two meters per second this is one meters per second and now we have to find a way to somehow Well, we have our opposite side and are adjacent side, so we can just use ah tangent to find her angle. So, yo, if you're trying to find an angle, you have to use 10 to the negative. One of this will be one meter per second, which is our opposite side, divided by our red side, which is two meters per second. So we're Sita. Our angle ultimately comes out to be 27 degrees after you plug this in your calculator. So whatever, he was helpful. Thanks for watching.

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