 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Motion Along a Line - Problem 2

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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We're talking about motion in a straight line. Here I have a problem that says; the position of an object is given by the parametric equations, x equals 2 minus 3t, and y equals -1 plus 4t. The first part asks me to find the velocity, and speed of the object.

Now I can find the velocity, if I write the equation of the object's motion in vector form. So let me take these parametric equations and put them in vector form. Remember that x and y, these are the components of the position.

So the vector for position equals the initial position; 2, -1 plus t times the velocity. That's going to be -3, and 4. Let's double check if this works. X equals 2 minus 3t, that's right. Y equals -1 plus 40, that's right also.

Now what you should know about this vector equation, is that this is exactly the velocity. Once you know that the velocity is -3, 4, well half of the problem is solved. Now I have to find the speed. Speed is the magnitude of velocity. So speed is the absolute value. That's the square root of -3² plus 4²; 9 plus 16, and that's 25. Square root of 25 is 5. So the velocity is -3, 4, and the speed is 5.

Now part b says; find the objects displacement from t equals 1 to t equals 6, and the distance travelled. Displacement is the change in position. So I'm going to need to know the position of the object at t equals 1, and t equals 6. So let me calculate that, t equals 1 first.

I'm going to use this position equation. So I have xy equals 2, -1 plus t. In this case 1 times -3, 4 velocity. So this is going to be 2 plus 1, times -3, 2 minus 3, -1, and -1 plus 1 times 4, -1 plus 4, 3. So that's the position at t equals 1. What about at t equals 6? Same equation, different time value. X,y equals 2,-1 plus, and now it's going to be 6, this is just t times -3, 4. That's 2, -1 plus, 6 times -3 is -18. 6 times 4, 24. 2 plus -18, is -16. -1 plus 24 is 23. So this is the position at t equals 6.

Displacement; remember that's the change in position, and specifically it's the position minus the old position. So it's -16, 23 minus the old position -1,3. So we subtract component-wise. -16 minus -1, -16 plus 1, 15. Then 23 minus 3, 20. So the displacement is -15, 20, and that means the particle is 15 units to left let's say, and 20 units above where it started.

What about the distance travelled? Distance travelled, as long as the object doesn't change direction, distance travelled is the magnitude of displacement. So the distance travelled is the magnitude of -15, 20. That's the square root of -15², 225 and 20², 400. That's the square root of 625 which is 25. The object has ended up 15 to the left, and 20 above where it started, which is 25 units away.