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Physics Chp 3. Trig Functions cos (θ) = x/h or =a/h sin(θ) =y/h or =o/h tan(θ) =y/x or =o/a soh cah toa. H or R R is the resultant. y. θ. x. scalars vs vectors vectors have a value and direction
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Physics Chp 3 Trig Functions cos(θ) = x/h or =a/h sin(θ) =y/h or =o/h tan(θ) =y/x or =o/a sohcahtoa
scalars vsvectors vectors have a value and direction resultant is the combination of vectors. We can add parallel vectors and we can break a vector into x and y components then adding them and calculating the resultant.
R2 = x2 +y2R = • tan(θ) = y/x θ= • If x = 3 and y = 2 find R
R2 = 32 +22R = • tan(θ) = 2/3 θ= tan-1(2/3) =
What about this? 8 R =? R is the resultant 11 θ
Still R2 = 112 + 82 R = tan(θ) = 8/11 θ= tan-1(8/11) =
We use the same equations for 2 dimensions as we did for one. Just label them x or y vx = vox +axΔt Δx = 1/2(vo + vx)Δt Δx = voxΔt + 1/2ax(Δt)2 vx2 = vox2 + 2axΔx vy = voy +ayΔt Δy = 1/2(vo + vy)Δt Δy = voyΔt + 1/2ay(Δt)2 vy2 = voy2 + 2ayΔy
Ex How long does it take a ball kicked off a 45 m tall building to hit the ground? • voy= 0 y= -45m ay= -9.8 m/s2 t = ?
Δy = voyΔt + 1/2ay(Δt)2 -45m = 0 + ½(-9.8m/s2)(Δt)2 t =
How far away from the building would the ball hit if you kicked it with a velocity of 5.4 m/s horizontally? vox = 5.4 m/s ∆t = 3.0s ax = 0 ∆x =?
Δx = 1/2(vo + vx)Δt OR Δx = voxΔt + 1/2ax(Δt)2 Δx = 5.4 m/s (3.0s) + ½ (0)(3.0s)2
EX. If a dog jumps from a dock with an initial velocity of 3.4 m/s at 15o above the horizontal, how high does the dog go? vo = 3.4 m/s θ = 15o ∆y= ?
vox = 3.4m/s cos15o = 3.3 m/s ax = 0 voy = 3.4m/s sin15o = 0.88 m/s ay = - 9.81 m/s2 How can we find the height?
vfy = 0 ! vy2 = voy2 + 2ayΔy O = (0.88 m/s)2 + 2(-9.81 m/s2) Δy Δy = 0.039 m
Relative Velocity If you are watching a person walk on the moving walkway and they are seem to be moving 5 mi/hr but you know they can only do 3 mi/hr, why is there a difference? What we perceive is relative to us or the situation. vp + vw = vobs.
This is also apparent when you approach an intersection and you are going 35 mi/hr while they are going 45 mi/hr perpendicular to you. You perceive the velocity to be the resultant? What is it and at what angle ?
v = ((35mi/hr)2 + (45 mi/hr)2 )1/2 v = 57 mi/hr θ = tan-1 (45/35) = 52o to the side of you.