A rocket-powered hockey puck moves on a horizontal frictionless t...

Question

A rocket-powered hockey puck moves on a horizontal frictionless table. Figure EX4.7 shows graphs of and , the x- and y-components of the puck's velocity. The puck starts at the origin. What is the magnitude of the puck's acceleration at t = 5s?

Accepted Answer

Hey, everyone in this problem, a radio controlled car with slicks coasts along a smooth tabletop. The motion is two dimensional in the X Y plane. And we're given the graph that show the X and Y components of its velocity. We're told to assume that the car's initial position is at the origin. And we're asked to find the car's acceleration at T equals 10 seconds. OK. So we're given two graphs here, both show a linear relationship. The first graph has the time in seconds on the X axis and the X component of the velocity in meters per second on the Y axis. And the second graph is very similar. But instead of the X component and has the Y component, we have four answer choices. Option A 7.55, option B 3.23, option C 1.4 and option D 4.48. And those are all in meters per second squared. So we're given a graph of velocity, we're asked to find the acceleration. So let's think about the relationship between the two. OK. And recall that the acceleration is related to the velocity through the derivative. So the acceleration is equal to the derivative of velocity with respect to time. So we have D V divided by D T and that's our acceleration. Now, when we're talking about a graph of the velocity instead of an equation, what's the derivative? Right? How do we represent the derivative? Oh It's just the slope. So the acceleration is just going to be the slope of the velocity time, Kirk. All right. So let's start in the X direction we have that the acceleration in the X direction is going to be equal to the derivative D V X by D T, which is just the slope of the V X curve. Now, when we're talking about slope, we have just a linear relationship here. We have a straight line. So we can think of this as the rise, which is gonna be the change in V X delta V X divided by the run, which is a change in time delta T OK. This is just rise over run. We know how to do that. Now, for the change in the X component of the velocity we're interested in the acceleration at 10 seconds. But again, because this is a straight line, we can choose any points we want in order to figure out what this acceleration is. And because anywhere around along this line is gonna give us the same slope, so let's choose the two end points. And those are gonna be easy to pick out the information from. We're gonna choose the two end points of our shown graph. Now the velocity, the X component at the left end point in the initial velocity is going to be negative eight m per second. And at the end it's 12 m per second. So the change is gonna be 12 m per second. That final velocity minus negative eight m per second, that initial velocity. And we're gonna divide this by the change in time that the final time is seconds and the initial time is zero seconds. So this is divided by 18 seconds minus zero seconds. If we simplify, you have that the acceleration is going to be equal to m per second divided by 18 seconds. The unit of seconds is going to multiply together meters per second per second, gives us meters per second squared. And if we simplify K dividing the numerator and denominator by two, we get that this is equal to 10 divided by nine m per second squared. OK? So we have the X component of our acceleration. We're gonna put a box around that. So we don't lose track of it. And now I wanna flip over to the Y direction and do the same thing. So the Y component of the acceleration A Y gonna be equal to the derivative of the Y component of the velocity with respect to time. This is just going to be the slope of R V Y. And again, that's gonna be the rise over run or the change and the velocity divided by the change in time, looking at our graph per V Y again, this is a straight line so we can choose any points. The slope is going to be the same. So let's choose the end point. We have a point at 00. And on the other end, we have a point at 14, 12, we're gonna choose those two points because they're easy to read off of that graph. The change in the Y velocity is therefore going to be 12 m per second minus zero m per second and the change in time is gonna be 14 seconds minus zero seconds. So going down to our equation and writing this out delta V Y 12 m per second minus zero m per second divided by delta T seconds minus zero seconds. OK? Simplifying gives us 12 m per second divided by seconds. And again, we can divide the numerator and denominator by two to simplify. And we get that this is equal to 67 and our unit is meters per second squared. OK? So now we have our Y acceleration as well as our X acceleration. Now, remember this question is asking us for the acceleration at T equals seconds. OK. So it wants the combined acceleration. It does not want the X component and the Y component given separately. So let's go ahead and draw a little diagram. To see how we can figure this out. Now, we have our X component of our acceleration that's positive. We're gonna have that pointing to the right, that's A X and we have the Y component that's also positive. We're gonna draw that pointing up A Y and the total acceleration, we're gonna add these tip to tail. And we can see that that's just going to be the hypotenuse of the triangle that we make here. That's our acceleration a, the hypotenuse of that triangle. And we can find that using Pythagorean theorem, OK. Recall that A squared is going to be equal to A X squared plus A Y squared according to Pythagorean theorem. OK. Substituting in our values we have that A squared is equal to 10, 9 meters per second squared, all squared plus six divided by seven m per second squared, all squared. If we simplify, we have the A squared is equal to approximately 1.969, 26 m squared 1st 2nd to the exponent four. And when we take the square root, we get that the acceleration is going to be equal to positive or negative 1.403 m per second squared. Now this is important when we take the square root, we always have to remember that we get the positive and the negative root. But we have to interpret which one is correct based on the physics. OK. In this problem, we have a positive X component and a positive Y component for our acceleration, the slope of these curves is positive so the acceleration should be positive. So we're gonna take the positive route we have that the acceleration is equal to 1.403 m per second squared. And that is exactly what this problem was asking for. If we go up to our answer choices. OK. And we round to two decimal places. You can see that the correct answer is option C 1.4 m per second squared. Thanks everyone for watching. I hope this video helped see you in the next one.

A rocket-powered hockey puck moves on a horizontal frictionless table. Figure EX4.7 shows graphs of and , the x- and y-components of the puck's velocity. The puck starts at the origin. What is the magnitude of the puck's acceleration at t = 5s?

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