PHY231 lecture3

# PHY231 lecture3 - Lecture 3 Some points of verticle motion...

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Lecture 3 Some points of verticlemotion: t = = downwards m/s 9.8 g 2 () t v v y y y gt v v o + = Δ 1 2 1 0 0 y g v v gt t v y Δ = = Δ 2 2 2 0 2 2 0 Here we assumed t 0 =0 and the y axis to be vertical. Vectors and components dditi f t Addition of vectors

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Graphical determination of displacement v v (m/s) t=1,v=2 2 (m/s) t=1,v=2 2 0 0 t (s) 0 t (s) 0 Q 1. 2. (m) area(m) 1)What is the displacement covered in 1 second? 2)What is the area indicated by ? Δ x (m) area(m) a) 1. 1. b) 1. 2. c) 2. 1. d) 2. 2. he area under the - rve is equal to Hint: Use Δ x=v 0 t+1/2at 2 or Δ x=(v 0 t+vt)/2 The area under the v t curve is equal to the displacement of the object! Note unit of area: area=2m/s*1s=2m
Example Imagine a car moving starting at rest at x=0 and having the following acceleration as a function of time. What is the velocity and position at 2 s and at 4 s? Can use area to get v(2) and v(4) v(2)=a Δ t=(1m/s 2 ) 2s=2m/s (4)=v(2)+ =2m/s+(2m/s 2 s v(4) v(2)+ a Δ t 2m/s+(2m/s ) 2s v(4)=6 m/s x by more obvious” route x=x 0 +v 0 t+1/2at 2 x(2)=0+0 (2s)+1/2 (1m/s 2 )(2s) 2 =2m x(4)=2m+2m/s 2s+1/2(2m/s 2 )(2s) 2 x(4)=2m+4m+4m=10m x by “less obvious” route se x(t)=x (v v(t)) 2 to get x(t) use x(t) x 0 +(v 0 +v(t)) t/2 to get x(t) x(2)=0+(0+2m/s) (2s)/2=2m x(4)=2m +(2m/s+6m/s) (2s)/2=10m

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Motion under Earth’s gravitational attraction Near the Earth’s surface, all objects are accelerated by gravity downwards with an acceleration of g=9.8 m/s 2 . = downwards m/s 9.8 g 2 () t v v y y y gt v v o + = Δ = 1 0 0 gt t v y = Δ 2 1 2 2 2 2 0 y g v v Δ = 2 0 A timed drop can provide ne measurement of g. one measurement of g.
Example 1 At t 0 =0 Lucy throws a ball straight upwards with a velocity of 30 m/s.

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## This note was uploaded on 05/03/2011 for the course PHYS 231 taught by Professor Nagy during the Fall '11 term at Michigan State University.

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PHY231 lecture3 - Lecture 3 Some points of verticle motion...

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