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Warm Up 1. A Child is pushing a shopping cart down an aisle at a speed of 1.5 m/s. How long will it take this child to push the cart down an aisle with.

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Presentation on theme: "Warm Up 1. A Child is pushing a shopping cart down an aisle at a speed of 1.5 m/s. How long will it take this child to push the cart down an aisle with."β€” Presentation transcript:

1 Warm Up 1. A Child is pushing a shopping cart down an aisle at a speed of m/s. How long will it take this child to push the cart down an aisle with a length of 9.3 meters? 2. A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system, and comes to rest 5.5 s later. Find the distance that the car travels during braking. Answers: s m

2 Vectors AP Physics 2

3 Definitions Scalar: A quantity with only a magnitude.
Vector: A quantity with a magnitude and direction. Vector Sum: The sum of two or more vectors. Vector Component: The projection of a vector along one axis.

4 Scalars and Vectors Scalar quantities: Temperature, pressure, energy, speed, wavelength, frequency, volume. Vector quantities: Displacement, acceleration, velocity, magnetic field strength. (Vectors tell nothing about the path, only the starting and ending points)

5 Vector Notation 𝑐 = π‘Ž + 𝑏 𝑏 π‘Ž 𝑐

6 Vector Addition rules 𝑏 π‘Ž 𝑐 𝑑 π‘Ž βˆ’ π‘Ž π‘Ž 𝑐 𝑑 𝑏 π‘Ž + 𝑏 = 𝑏 + π‘Ž
( π‘Ž + 𝑏 )+ 𝑐 = π‘Ž +( 𝑏 + 𝑐 ) 𝑏 π‘Ž 𝑐 𝑑 βˆ’ π‘Ž π‘Ž π‘Ž 𝑐 𝑑 𝑏

7 Components of Vectors π‘Ž= π‘Ž π‘₯ 2 + π‘Ž 𝑦 2 π‘Ž π‘Ž 𝑦 tanπœƒ= π‘Ž 𝑦 π‘Ž π‘₯ π‘Ž π‘₯
Vector Component: The projection of a vector along one axis. π‘Ž= π‘Ž π‘₯ 2 + π‘Ž 𝑦 2 π‘Ž π‘Ž 𝑦 π‘Ž 𝑦 =π‘Ž sinπœƒ πœƒ= 30⁰ tanπœƒ= π‘Ž 𝑦 π‘Ž π‘₯ π‘Ž π‘₯ π‘Ž π‘₯ =π‘Ž cosπœƒ

8 Adding Vectors 𝑏 =7π‘š π‘Ž =5π‘š 𝑐 =?
πœƒ 3 =? 𝑐 = π‘Ž + 𝑏 π‘Ž =5π‘š, πœƒ 1 =80⁰ 𝑏 =7π‘š, πœƒ 2 =75⁰, πœƒ 3 =-15 ⁰ 𝑏 =7π‘š πœƒ 2 =75⁰ π‘Ž =5π‘š π‘Ž π‘₯ =5cos80Β°=.868π‘š 𝑏 π‘₯ =7cos(βˆ’15Β°)=6.761π‘š πœƒ 1 =80⁰ 𝑐 =? π‘Ž 𝑦 =5sin80Β°=4.924π‘š 𝑏 𝑦 =7sin βˆ’15Β° =βˆ’1.812π‘š 𝑐 π‘₯ =.868π‘š+6.761π‘š=7.63π‘š 𝑐 𝑦 =4.924π‘š+ βˆ’1.812π‘š =3.11π‘š 𝑐= 𝑐 π‘₯ 2 + 𝑐 𝑦 2 = =8.24π‘š tanπœƒ= 𝑐 𝑦 𝑐 π‘₯ , πœƒ=tan-1 𝑐 𝑦 𝑐 π‘₯ =tanβˆ’ =22.2Β°

9 Practice Chapter 3, Problems 1, 3, 7, 9, 11, 15.


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