 ###### 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.

##### Thank you for watching the video.

To unlock all 5,300 videos, start your free trial.

# Motion Along a Line - Problem 1

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.

Share

We're talking about motion along a line. Here is another problem. A particle is moving with constant velocity, v, equals and in component form 8,-6. At t equals 2, the particle is at the point 21,-4. Where was the particle at t equals 0?

Well, let's start by observing that if the particle has constant velocity, it's got to move in a straight line. An equation for it's motion would be r equals r0 plus the scalar t, times velocity. We don't know the initial position. We don't know this vector, but we do know that when t equals 2, the particle is at 21, -4. Let me plug that in.

Now the position vector for that point would be 21, -4 equals r, 0 and when t equals 2, this happens. 2 times the velocity vector 8, -6. Let me distribute that 2 over the velocity vector. I get 2 times 8, 16 and 2 times -6, -12. Then over here I've got 21, this is -4.

Let me subtract this vector from both sides. I'm trying to solve for our zero. So when I subtract remember I subtract component y, so I get 21 minus 16, 5. -4 minus -12, -4 plus 12, 8. So that's my r so zero. What that means is that at t equals 0, the particle was at the point 5, 8. The particle was at 5, 8.

Now it says find vector, and parametric equations for the position of the particle. We're pretty much there, as far as vector equations go. Let's take a look. This is our vector equation. We now know our 0 is 5, 8 and we knew the velocity was 8, -6. We get r equals 5, 8, plus t times 8, -6 which can also be written x, y equals 5, 8, plus t times 8, -6. That's a vector equation.

What about parametric equations? Well, you can always take a vector equation apart to get parametric equations. Take a look at all the first components, the horizontal components. X equals 5 plus t times 8, that's one of them. X equals 5 plus 8t. The other y equals 8, plus t times -6. 8 minus 60. You can always get you parametric equations by looking at all of the first components, and all of the second components.

Finally, find the speed of the particle. What you should know about speed is that speed is the magnitude of velocity. This is our velocity vector right here. So speed is the magnitude of velocity. So it's the magnitude of the vector 8, -6. That's the square root of 8², or 64 plus -6² is 36. That adds up to 100, square root of 100 is 10. So the speed of the object is 10.

Remember the difference between speed and velocity. Velocity gives you more information, speed is just magnitude. Velocity gives you magnitude, and direction. Also the two forms of this equation; the vector form and the parametric equations form. Both of these describe the same path.

© 2019 Brightstorm, Inc. All Rights Reserved.