51. Frictionless cart A small frictionless cart, attached to t...

Question

51. Frictionless cart A small frictionless cart, attached to the wall by a spring, is pulled 10 cm from its rest position and released at time t = 0 to roll back and forth for 4 sec. Its position at time t is s = 10 cos πt.

a. What is the cart's maximum speed? When is the cart moving that fast? Where is it then? What is the magnitude of the acceleration then?

Accepted Answer

Welcome back, everyone. In this problem, two trains, train A and B, are initially 30 miles apart with train A to the west of train B. Train A moves east at a speed of 20 MPH, while train B moves south at a speed of 15 MPH. Determine the function that represents the distance D between the two trains as a function of time T, where T equals 0 at the moment they start moving. Now for us to best find the distance function, it would be a good idea for us to set up this coordinate system. So let's draw uh some axis here, OK. So here we have our axis. And based on our problem statement, OK, let's place train B at the origin. OK? So this is where train B is at. A 00. Now what do we know? Well, we know that train A is 30 miles to the west of train B. So on our coordinate system, we could put it at -300, OK. So here, this is where we have train A. OK. And for reference, let's just put in our north, south, east and west markers for our axes. Now what else do we know? Well, remember that in our problem we're told that train A moves east at a speed of 20 MPH while train B moves south at a speed of 15 MPH. So we can define the position of each train as a function of time because for train A, OK. We can put it here on our diagram moving east along the x axis and thus the position of train A at the time of T is gonna be equal to, let me put that in red here, uh negative 30 + 20 T. For our X coordinate and 0 for or Y coordinate, OK. That would represent a moving at a speed of 20 MPH. On the other hand, remember we're told that train B moves south at 15 MPH. So if we can represent that on our grid, OK, 15 MPH, its position function or it's, is gonna be 0 and -15T, OK? Because it's moving along the negative Y axis, 15 MPH south. So here, now that we have all the information we need, how can we use that to find the distance D between the two trains as a function of time, T? Well, let's let D of D be the distance between the trains at a time T, OK? Recall that the distance function basically tells us that we are going to find the square root of the distance of the difference between the X coordinates squared, OK, plus the difference between the Y coordinates squared. So in this case applied to our time or D acting as a function of time. That means a day of tea. It's gonna be equal to the square root of the difference between X A of T minus XB of T. All squared, OK. Plus, the difference between Y A of T minus YB of T all squared respectively, OK? So now if we substitute in our coordinates, we know that X A of T, the X value of our of of train A as a function of time. equals -30 + 2 + 20T. We already found that here, OK. And the Y value as a function of time is equal to 0, OK, because it's going along our X-axis. While the X value of chain B as a function of time is equal to 0, and the Y value of chain B as a function of time equals -15T. So now if we substitute those values into the distance function. Then it's gonna be equal to the square root of -30 + 20 T minus 0 squared. Plus 0 minus -15t all squared. And now in this case, if we simplify, OK, this is gonna be equal to the square root of -30 plus 20 T all squared. Plus 15 and T squared. And when we expand that, OK, then we could write this as the square root of 400 square minus 1,200T plus 900. OK. That is what -30 + 20 T2 would give us plus, OK. 225 T squared. And now if we simplify, OK, we can combine our T2 terms, 400 + 225 equals 625. So that's 625 T2 minus 1,200T plus 900. So this is going to be our distance function in terms of time T, OK, in miles. Fort greater than or equal to 0. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Simple Harmonic Motion

Velocity in SHM

Acceleration in SHM

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