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Chapter 3: Vectors and Parabolic Motion

Chapter 3: Vectors and Parabolic Motion. Vectors. Remember, a vector is something that has both a magnitude and a direction . Magnitude = how long the vector is Direction = where the vector points

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Chapter 3: Vectors and Parabolic Motion

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  1. Chapter 3: Vectors and Parabolic Motion

  2. Vectors • Remember, a vector is something that has both a magnitude and a direction. • Magnitude = how long the vector is • Direction = where the vector points • Ex = A boat that has a velocity of 35mi/hr at a heading of N 30o E has a magnitude of 35mi/hr and a direction of N 30o E.

  3. A word about heading • The A blah0 B notation is new to most of us. • The way to read it is this: Start at the first direction and swing out toward the second direction by the given degree. • As an example I will draw W 30oS for you.

  4. Some practice • Draw the following vectors • N 45o E • E 30o S • W 60o N

  5. Components • The easiest way to add vectors together is by adding their components. • A component is a piece. Most, if not all, of the vectors we will work with can be split into 2 pieces, an X piece and a Y piece. • X piece = how many units sideways the vector is . • Y piece = how many units up and down the vector is.

  6. Finding your pieces • If your vector lies on one axis it is easy to find your components, one is 0 and the other is all of it. • But if your vector lies between two axes, use the following 2 equations • Vy = V sinθ • Vx = V cosθ • Where θ is the angle the vector makes with the x axis and V is magnitude of the vector

  7. Example • A quarterback throws a long pass a distance of 35yds at a direction of N 30o W. How many yards north does the receiver have to run to meet the ball?

  8. Solution • Draw a picture • Our west component is x, and our north component is y, so we need to find our y piece. • Vy = V sinθ • Vy = 35yds sin 60 = 30.3 yards

  9. Your Turn • A pool ball is shot 3feet in a direction of E 30o N. Find the length of the x component.

  10. What you can do with your pieces • If you know both components of a vector you can find 2 things: • How long the vector is using √(Vx2 + Vy2) • Where the vector points using tanθ = Vx / Vy and using tan -1 to find θ.

  11. Adding 2 vectors (or more) • To add two vectors break each vector into its 2 components giving you vx1, vy1, vx2, and vy2 • The resulting vector, called the resultant vector (catchy name) or R. • Rx = vx1+ vx2 • Ry = vy1+ vy2 • Use the 2 equations on the last slide to find magnitude and direction.

  12. Example • A mailman leaves the post office and drives 22.0 km north. Then he drives E 60o S for 47.0km. What is his displacement from the post office? (aka how far is he from the post office?)

  13. Practice Problem • A plane flies 600 km at a direction of N 30o E, and turns and flies 450 km at a direction of N 600 E. Find the displacement of the plane. (this means both magnitude and direction)

  14. Projectile Motion • Imagine a car driving off a cliff. • The only thing acting on it as it sails through the air is gravity. • Of course it has a sideways velocity before driving off the cliff and nothing is acting on it sideways to change its speed to it will continue to move sideways as well as fall.

  15. Parabola • This motion of moving sideways while falling down creates a curve, or parabola. • Brain storm: What factors would affect the shape of the parabola?

  16. Some tools • As stated earlier, there is nothing happening in the horizontal direction to change the speed of the object in that direction. In other words a = 0 and the velocity in the x direction, vx, is constant. This changes 2 of our kinematic equations: • v = v0 + at becomes vx = vx0 and • x = x0 + v0t + ½ at2 becomes x = x0 + vx0t

  17. Meanwhile, in the vertical direction • The only thing acting in the vertical is gravity, so a = g = -9.8m/s2 (or -32ft/s2 in American units) • This gives us the following: • vy = vy0 – gt • y = y0 + vy0t – ½ gt2 • vy2 = vy02 – 2gy

  18. The power of math • These 5 equations can tell us very important things about an object in projectile motion including the time it spends in the air, where it will land, how high it will get, and how fast it was going at the start. • This is used everywhere from weather balloons to cruise missiles.

  19. Example • A stunt driver on a motorcycle drives off a 50.0m high cliff. How fast must the motorcycle leave the top of the cliff if it is to land in front of the cameras 90.0m away from the cliff?

  20. Solution Part 1 • y0 = 0m • y = -50.0m • x0 = 0m • x = 90.0m • g = -9.8m/s2 • v = ? • t = ?

  21. It’s about time • Because vx = vx0 we would know v if we knew t by using x = vt • So can we find t? • Yes! We can use y = y0 + vy0t – ½ gt2 • y = -1/2gt2 gives us t = 3.19s • Now we can use x = vt, which gives v = 28.2m/s

  22. Range • Before we examine flying fruit, there is one more equation you will need to know. • R = x = vx0t = • Where θ is the angle the object is launched at from the ground. • Protip: when θ = 45the above equations becomes v02 / g

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