1.5 Frames of Reference and Relative Velocity

A frame of reference is a coordinate system relative to the motion of an object. Keyword: RELATIVE Relative Velocity: the velocity relative to a specific frame of reference

Example #1: Consider a person in a train. The train is moving 30.0 m/s [S]. a) If the person is sitting on the train, what is the person’s velocity relative to the train? b) If the person is sitting on the train, what is the person’s velocity relative to the ground?

c) If the person is walking on the train at 1 c) If the person is walking on the train at 1.0 m/s [S], what is the person’s velocity relative to the ground? d) If the person is walking on the train at 1.0 m/s [N], what is the person’s velocity relative to the ground?

e) If the person is running on the train at 8 e) If the person is running on the train at 8.0 m/s [W], what is the person’s velocity relative to the ground?

Example #1 Solution a) The person’s velocity relative to the train is zero, since sitting, thus not moving relative to the train.

b) vpg = velocity of the person relative to the ground → b) vpg = velocity of the person relative to the ground vpt = velocity of the person relative to the train vtg = velocity of the train relative to the ground

The person’s velocity relative to the ground is 30.0 m/s [S]. → → → vpg = vpt + vtg = 0 m/s +(–30.0 m/s) = – 30.0 m/s = 30.0 m/s [S] The person’s velocity relative to the ground is 30.0 m/s [S].

The person’s velocity relative to the ground is 31.0 m/s [S]. → → → c) vpg = vpt + vtg = –1.0 m/s +(–30.0 m/s) = – 31.0 m/s = 31.0 m/s [S] The person’s velocity relative to the ground is 31.0 m/s [S].

The person’s velocity relative to the ground is 29.0 m/s [S]. → → → d) vpg = vpt + vtg = +1.0 m/s +(–30.0 m/s) = – 29.0 m/s = 29.0 m/s [S] The person’s velocity relative to the ground is 29.0 m/s [S].

e) Since this question is 2D, the person’s velocity relative to the ground can NOT be calculated simply by adding or subtracting the numbers. The solution is to use the components method or using the Sine Law and Cosine Law. → → → vpg = vpt + vtg

vpg2 = vpt2 + vtg2 vpg2 = (8. 0 m/s)2 + (30 vpg2 = vpt2 + vtg2 vpg2 = (8.0 m/s)2 + (30.0 m/s)2 vpg2 = 64 m2/s2 + 900 m2/s2 vpg2 = 964 m2/s2 vpg = 31.0 m/s

tan θ = vpt vtg tan θ = 8.0 m/s 300 m/s tan θ = 0.2667 tan θ = o/a tan θ = vpt vtg tan θ = 8.0 m/s 300 m/s tan θ = 0.2667 θ = tan-1(0.2667) = 14.9º

Thus the direct is [S 14. 9º W] Thus the direct is [S 14.9º W]. There the person’s velocity relative to the ground is 31.0 m/s [S 14.9º W].

Example # 2: A plane is flying due west at a speed of 200 km/h into a 50 km/h wind blowing towards the east. What is the plane’s velocity relative to the ground?

vag = velocity of the air relative to the ground (wind velocity) Solution → vpg = velocity of the plane relative to the ground vpa = velocity of the plane relative to the air vag = velocity of the air relative to the ground (wind velocity) → → → vpg = vpa + vag

The plane’s velocity relative to the ground is 150 km/h [W]. → → → vpg = vpa + vag = (–200 km/h)+50 km/h = – 150 km/h = 150 km/h [W] The plane’s velocity relative to the ground is 150 km/h [W].

Example # 3: A plane is flying at 200 km/h [N 40 º E] relative to the air. There is a wind blowing due east at 50km/h. What is the plane’s ground velocity?

Solution

vpg2 = vpa2 + vag2 – 2vpavagcos = (200 km/h)2 + (50 km/h)2 – 2(200 km/h)(50 km/h) cos(40º + 90º) = 40 000 km2/h2 + 2500 km2/h2 + 12856 km2/h2 = 55356 km2/h2 vpg = 235.28 km/h

a b sin θ = sin 130º vag vpg sin θ = vag (sin 130º) vpg sin A = sin B a b sin θ = sin 130º vag vpg sin θ = vag (sin 130º) vpg sin θ = (50 km/h) (sin 130º) 235.28 km/h

sin θ = (50 km/h) (sin 130º) 235. 28 km/h = 0. 1628 θ = sin-1(0

 = 40º + 9. 4º = 49. 4º Therefore the plane’s ground velocity is 2  = 40º + 9.4º = 49.4º Therefore the plane’s ground velocity is 2.4 X 102 km/h [N 49.4º E].

Example # 4: From Town A to Town B are on a straight river Example # 4: From Town A to Town B are on a straight river. Town B is a 150 km down stream of Town A. A boat travelling with its maximum speed can travel from “A to B” in 5.0 h. The return trip at maximum speed takes 7.5 h.

a) What is the speed of the current a) What is the speed of the current? b) What is the maximum speed of the boat? c) How long would it take to drift from Town A to Town B?

Solution

vbg = velocity of the boat relative to the ground → vbg = velocity of the boat relative to the ground vbw = velocity of the boat relative to the water → → → vbg = vbw + vwg

a) v = Δd Δt (since constant velocity)

Boat travelling from “A to B” vbg(AtoB) = 150 km 5.0 h = 30 km/h

→ → → vbg(AtoB) = vbw + vwg Let downstream be “+” and upstream be “–” vbg(AtoB) = + vbw + vwg {Equation #1} 30 km/h = + vbw + vwg

Boat travelling from “B to A” vbg(BtoA) = 150 km 7.5h = 20 km/h

Let downstream be “+” and upstream be “–” –vbg(BtoA) = –vbw +vwg → → → vbg(BtoA)= vbw + vwg Let downstream be “+” and upstream be “–” –vbg(BtoA) = –vbw +vwg {Equation #2} –20 km/h = –vbw +vwg

{Equation #1} + {Equation #2} 30 km/h–20 km/h = +vbw +vwg –vbw +vwg 10 km/h = 2vwg vwg = 10 km/h 2 vwg = 5.0 km/h Therefore the speed of the current is 5.0 km/h.

b) {Equation #1} – {Equation #2} 30 km/h –(–20 km/h) = + vbw + vwg –(– vbw+ vwg) 30 km/h +20 km/h = + vbw + vwg + vbw – vwg 50 km/h = 2vbw 2vbw = 50 km/h vbw = 50 km/h 2

2vbw = 50 km/h vbw = 50 km/h 2 vbw = 25 km/h Therefore the maximum speed of the boat is 25 km/h.

c) The boat drifting from Town A to Town B v = Δd Δt vΔt = Δd Δt = Δd v

Δt = Δd v = 150 km 5.0 km/h = 30 h Therefore it would take the boat 30 h to drift from Town A to Town B.

Practice Question #1: A plane is traveling due North with an airspeed of 200 km/h. There is a 50 km/h wind blowing due east. What is the velocity of the plane relative to the ground?

Practice Question #2: A plane wants to head due North, unfortunately there is a wind blowing due west at 50 km/h. The plane has an air speed of 200 km/h. What should the direction of the plane be flying relative to the air? What is the plane’s velocity relative to the ground?

1.5 Practice Questions Page 56 Questions 1-5 Page 57 Questions 5-6 From Nelson 12 Physics (pdf text) Page 56 Questions 1-5 Page 57 Questions 5-6 (pictures of question on next pg)