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. (b) How far from the origin is the puck at t = 5s?

Video transcript

Hi, everyone in this practice problem, we are being asked to calculate the discs position. After two seconds, we will have a small disc sliding on a very smooth horizontal surface. And the variations of the horizontal and vertical components of the discs velocity are represented in the figure below at time T equals to zero. The disc is at the origin at 00.0, comma zero. And we're being asked to calculates the discs position. After two seconds, we were given the at V X versus time and also P Y versus time figures or diagrams where V X is in centimeters per second. P Y is also in centimeters per second and T is in seconds for both figures in the figure of V X versus T, the velocity of V X stays constant at two centimeters per second from a point a time of zero till seven seconds. And for the P Y versus the graph, the uh V Y velocity in the Y direction uh started from 0.3 centimeters per second and decreases until zero centimeters per second at time equals to six seconds. So the options given for the position after two seconds are A are of two seconds equals to zero, I plus three J in centimeter B R of two seconds equal equals to one I plus one J in centimeters, C R of two seconds equals to four, I plus five J in centimeters and D R of two seconds equals to 12 I plus 18 J also in centimeters. So in order for us to calculate the disks position, we will model the disc as a point like particle under constant acceleration. So at the equals to zero seconds, the desk will equals or the desk will be at the origin of X knot equals zero centimeter and Y knot equals zero centimeter. So I'm just gonna write that down real quick. So at T equals zero seconds, X knot equals zero centimeter. And Y knot also equals to zero centimeters, we will apply the kinematic equation along the X N D Y directions. So first let's start with moving along the X direction. So along the X direction, we are going to focus on our V X versus the figure. So in calculus terms V X is the derivative of X. So V X equals to D X over D T. So rearranging this, we can get D X equals to V X multiplied by D T. We want to integrate, integrate both sites from 0 to 2 seconds. So um for the left side, we are going to integrate from at P equals to zero seconds. X knot is going to be at zero centimeters. So X knot, we're going to integrate from X knot till X of two seconds. D X equals to the integral. For the right side, we want to start integrating from T equals to zero seconds to T equals to two seconds of P X multiplied by D T just like. So, so for the left side, we know X knot is zero. So essentially this left side will be X of two seconds minus X knot. And that all equals two, the integral of starting from T equals to zero seconds to T equals to two seconds. F V X D T just like. So, so the integral from T equals to zero seconds to T equals to two seconds of V X TT will equals to the area under the V X of T curve up until the second seconds. So that area will actually correspond to this rectangle right here just like. So, so that area will correspond to the right side of our integral. So the integral from T equals to zero seconds to T equals to two seconds of P X D T will equals two, two cm per second minus zero, multiplied by two seconds minus zero seconds. And that will essentially give us a value of four centimeter. So from this, we can then uh input this value into our integral equation. So X of two S will minus X knot equals to four centimeters. So finally, the horizontal position of the disc at B equals two seconds is then going to equals X of two seconds equals to, we know X NOTT is zero centimeters. So equals to four centimeters. So the X position after two seconds is going to equals to four centimeters. Awesome. So now the second step is to find the vertical position now that we found our horizontal position. So the vertical position in the right direction direction uh is going to be given by pretty much the same thing. So P Y will then equals to D Y over D T. And after sum rearrangement, then D Y will equals to P Y multiplied by D T. We are interested in looking at point from 0 to 2 seconds. So essentially, we are going to integrate from 20.0 to 2 seconds. So the integral from Y not of two Y of two seconds, D Y will then equals to the integral of T from T equals to zero seconds to T equals to three second or two seconds. R V Y D T, the integral of the left side will then be Y of two seconds minus Y not equals to the integral of starting from T equals to zero seconds to T equals to two seconds of P Y D T. And similarly to the previous uh X direction, we can find the right side of the integral from the figure that we have or the figure that are given in our problem statement. So the area from 0 to 2 seconds of V Y versus D graph will be represented by this area right here. And the area can actually be separated into two different colors or two different parts. So the first part is going to actually be our rectangle, which is pretty much the same as, which is pretty much going to be the same as our V X versus the area. And the additional part is going to be the triangle shown in red oops. So we wanna calculate that, we want to calculate the service area under the V Y T curve up until the second seconds. So calculating that the right side of the integral will then equals to the integral from T equals to zero seconds to T equals to two seconds of P Y TT will equals two, the area of the rectangle first, which is two centimeters per seconds minus zero centimeters per seconds multiplied that by two seconds minus zero seconds Plus the area of the triangle, which is going to be three cm per second minus two centimeters per seconds multiplied that by two seconds minus zero seconds divided by two just like. So, so that is coming from uh the three centimeters minus two centimeters is going to be the base of the triangle or essentially the height and the two seconds minus zero seconds is going to be the base and multiply that by a half to calculate the area of the triangle that will give us a value of the integral from T equals to zero seconds to T equals to two seconds of V Y D T two B five centimeter just like. So OK. So at T equals two, so at T equals to two seconds, we can then find by substituting this value into our final integral equation. So Y E of two seconds minus Y not equals to five centimeter just like. So and we know that Y not is zero. So Y of two seconds equals to five centimeter. So at T equals two seconds, X will equals to four centimeter and Y will equals to five centimeters. So at P equals two seconds, the disc will be located or R of two seconds will be four, I plus five J sent the reader just like. So the four is coming from The X after two seconds just like so And the five is coming from the Y at the two seconds point. So that will be all for this particular practice problem where the answer of R or position at two seconds equals to four I plus five J centimeter that will correspond to option C in our answer choices. That'll be all for this particular practice problem. So option C is going to be the answer to this video. 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.