P2F11-Week5-Barwick

# P2F11-Week5-Barwick - P2 Week 5 Solving two equations...

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P2 Week 5 Solving two equations simultaneously Introduction to Vectors Addition and subtraction of Vectors Reading assignment 1.7, 1.8, 1.9

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Two equations for two unknowns l If you have N unknowns, you at least need N equations to solve them. l Simple example: consider two independent variables x and y. You are given l 2=x–y l 4=x+y l Solution for x is 3 and y is 1, which you can find (for example) by substituting y=x-2 (from the first equation) in the second.
Two equations for two unknowns l Another example: consider two independent variables x and y. You are given l 2=x–y l 8=x 2 –y 2 +2x l Solution for x is l A: 1 B: 2 C: 3 D: 4 l E: None of the above

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Two equations for two unknowns l Solving 1D and 2D equations of motion requires solving systems of equations with unknowns occurring as linear or quadratic terms. l When you have only two equations, the method of solving this set of equations is simple and requires substituting the relation between the two observables from one equation into the other.
1D equation of motion l Displacement d = 10 m l Acceleration a = –1 m/s 2 l Suppose we wish to find the initial and final velocities. Do you have enough information to solve this problem? A: Yes B: No

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1D equation of motion l Of the 4 kinematic equations, only 2 are independent l We derived the the final two from the first two. l In fact, given any 2 of the 4, we can derive the other two. l You need at least three of the following five for each constant acceleration motion: (t f -t i ), (x f -x i ), a, v f , v i
Find height of building (review) l Elizabeth and Kelsey do the following experiment to figure out the height of a building. Elizabeth drops a ball from the roof and at the same time Kelsey drops a ball from 5m below the roof. Kelsey’s ball strikes the ground 0.20s earlier than Elizabeths . What are the equations that describe the motion? Use g= -10 m/s 2 . Call tE , the time for ball from Elizabeth to strike the ground. Origin is located at Elizabeth. l A: dE= -5(tE)^2 ; dK=5-5(tK)^2 ; tK=tE+0.2, dK=dE l B: dE= -5(tE)^2 ; dK= -5-5(tK)^2 ; tK=tE+0.2, dK=dE l C: dE= -5(tE)^2 ; dK=5-5(tK)^2 ; tK=tE-0.2, dK=dE l D: dE= -5(tE)^2 ; dK=-5-5(tK)^2 ; tK=tE-0.2, dK=dE l E: dE= -5(tE)^2 ; dK=5-5(tK)^2 ; tK=tE+0.2, dK=dK

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Find height of building (review) l Elizabeth and Kelsey do the following experiment to figure out the height of a building. Elizabeth drops a ball from the roof and at the same time Kelsey drops a ball from 5m below the roof. Kelsey’s ball strikes the ground 0.20s earlier than Elizabeths. What is the height of the building? Use g= -10 m/s 2 . l A: 16 m l B: 22 m l C: 27 m l D: 31 m l E: 34 m
Moejoe Clicker problem l Moe needs an orange and an apple to make fruit punch. Joe two floors above throws an apple down from his window with some speed. At the same time, Joey one floor below her throws an orange upward with the same speed,v. Moe catches a fruit in each outstretched hands (palms up) at the same time. Assume: l (displacement of apple) = –2 × (displacement of orange) and that the apple travels 4m between Joe ` s and Moe ` s hands. g=10m/s 2 .

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## This note was uploaded on 12/12/2011 for the course PHYS 2 taught by Professor Staff during the Fall '08 term at UC Irvine.

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P2F11-Week5-Barwick - P2 Week 5 Solving two equations...

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