A rocket-powered hockey puck moves on a horizontal frictionless table. FIGURE EX4.6 shows graphs of and , the x- and y-components of the puck's velocity. The puck starts at the origin. (a)In which direction is the puck moving at t = 2s? Give your answer as an angle from the x-axis.

Video transcript

Hi, everyone in this practice problem, we're asked to determine the direction of the velocity factor of the ice skater with respect to the positive horizontal X direction at time T equals three seconds. There is the horizontal and vertical components are given in the two figures of the velocity of an ice skater performing in a horizontal frictionless area at T equals to zero seconds. The ice skater's initial position factor is equals to R not equals zero I plus zero J or essentially located at the origin. We're asked to determine the direction of the velocity factor of the ice skater. We respected a positive horizontal direction at a specific time of three equals three seconds. The first figure shown is showing P X or the X velocity component versus time. And the second figure shown is showing V Y or the Y velocity component versus time. Both the velocity are in meters per seconds and the time is in seconds. The V X versus time goes from 0.0 comma minus three to all the way straight uh throughout the duration shown in the figure. So the V X remains constant at minus three m per seconds while the V Y versus D, the V Y goes from 30.0, comma 62.6 comma zero. So the V Y is essentially decreasing from six m per second. All the way to zero m per second at time equals six seconds. The answer choices given are 30 degrees clockwise with respect to the positive X axis, 45 degrees clockwise with respect to the positive X axis, 30 degrees counterclockwise with respect to the positive X axis and 45 degrees counterclockwise with respect to the positive X axis. So using the data shown in both graphs, we want to first determine what the velocity factor is of the ice skater at time. Three equals T equals three seconds. So at time T equals equals three seconds, then V will equals two V X in the I component plus V Y in the J component. So at three P equals three seconds, the V X given in the figure is going to equals to minus three because it stays constant during the whole time at minus three m per second. So we have minus three I plus the V Y. So at three seconds, the V Y is going to be at three m per second. So this, the V Y is going to be three G and all of this is going to be in meters per second. So now that we have the velocity factor of the ice skater at time, T equals three seconds, the direction of the motion is going to be the same as the direction of the velocity factor. Let the be the direction of the velocity factor with respect to the horizontal direction or the positive the X axis direction. Therefore, the angle theta can actually be computed using this following equation right here, theta will equals to the, the tangent or the in 1st 10 10 to the power of minus one or the in first tangent of P Y over P X, we are letting theta to be um the direction of the velocity factor with respect to the horizontal direction or the positive XX access direction. So now we want to input the V Y and the V X value. So in first tangent or tangent to the power of minus one or V Y, which is three of V X which is minus three. So this will give us the inverse tangent of minus one, which will equals to minus 45 degrees because of the negative 45° sign right here. The direction of the velocity factor is then going to be 45° clockwise with respect to deposit horizontal direction. So degrees clockwise with respect 2 Positive Horizontal Direction. Essentially the way we know that it is going to be clockwise is true. The factor uh or the clockwise and anticlockwise or counterclockwise convention that we have um when we're going counterclockwise, counterclockwise is going to always be positive while we're really where, where we're going clockwise, it is going to be negative now because we have negative 45° as the data values, we know that it is going to be going clockwise with respect to the positive horizontal direction or positive access. And that will actually correspond to option B in our answer choices. So answer B is going to be the answer to our problem statement and that'll be all for this one. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and they'll be all for this one. Thank you.