Relative Velocities In Two Dimensions

Transcription

1 Relative Velocities In Two Dimensions The heading is the angle of the moving body the direction the object is pointing. The resultant velocity is typically the velocity of the object relative to the ground. An observer at rest on the ground would observe this motion. Ex. boat in a river animation Play with the current speed, the boat speed, and the boat's headings. For example, point the boat such that it has a heading vector in quadrant 1 with an eastward current. Then, increase the current's speed. What changes about the boat's resultant velocity? Now switch the boat's heading to a vector pointing in quadrant 2. What change's about the boats resultant velocity? And, what heading would the boat need to have it land due north of where it starts? Change the current's speed. How do you have to modify the points heading in order to maintain a resultant velocity due north?

2 Physics McGraw Hill Ryerson 2002 pp

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4 Relative Velocity (Perpendicular Cases) Ex. 1 A river flows from the west to the east. It is 750 m wide. Amanda and her friends are on the south shore of the river. Amanda can swim 2.0 m/s relative to the water. Amanda's heading is always north. The river's current is flowing 1.0 m/s [E] relative to the ground. a) What is Amanda's resultant velocity as witnessed by her friends on the shore? V A/W = 2.0 m/s [N] V W/G = 1.0 m/s [E] V A/G =? note VR = V A/G sketch VA/G = V A/W + V W/G V W/G OR V A/G VA/W N θ W V A/G N W α V W/G VA/W VA/G = V W/G + V A/W because order of vectors does not matter when adding... vector magnitude VA/G = VA/W 2 + VW/G 2 V A/G = (2.0 m/s) 2 + (1.0 m/s) 2 V A/G = (4.0 m 2 /s 2 ) + ( 1.0 m 2 /s 2 ) V A/G = 5.0 m 2 /s 2 V A/G = m/s vector direction tan θ = V W/G V A/W θ = tan m/s 2.0 m/ s θ = o θ = 26.6 o Final answer is V A/G = 2.2 m/s [N 26.6 o E] I could have worked with the second triangle but would have solved for α instead of θ. b) How much time does it take Amanda to swim across the river (assuming velocities remain constant)? Here, we must match the displacement to the velocity vector. They both must be in the same direction. Since I am given the north displacement (from south bank to north bank), I will use the north velocity. Δdy = 750 m [N] Vy = V A/W = 2.0 m/s [N] Δt =? From last year, assuming constant velocity for now to keep it simple, Vy = Δdy Δt Δt= Δdy Vy Δt = 750 m 2.0 m/s = 375 s = 3.8 x 10 2 s c) How far downstream would Amanda land? Must match the displacement to the velocity vector. They both must be in the same direction. Since I am calculating the easterly displacement, I will use the eastward velocity. Δdx =? Vx = V W/G = 1.0 m/s [E] Δt = 3.8 x 10 2 m/s The time moving east equals the time moving north. Vx = Δdx Δt Δdx = Vx Δt Δdx = (1.0 m/s) (375 m/s) = 375 m = 3.8 x 10 2 m [E]

5 Try the next one!

6 Ex. 2 A river flows from the west to the east. It is 750 m wide. Amanda and her friends are on the south shore of the river. Amanda can swim 2.0 m/s relative to the water. Amanda's heading is always north. The river's current is flowing 5.0 m/s [E] relative to the ground. a) What is Amanda's resultant velocity as witnessed by her friends on the shore? b) How much time does it take Amanda to swim across the river (assuming velocities remain constant)? c) How far downstream would Amanda land?

7 Answers on next page!

8 Ex. 2 A river flows from the west to the east. It is 750 m wide. Amanda and her friends are on the south shore of the river. Amanda can swim 2.0 m/s relative to the water. Amanda's heading is always north. The river's current is flowing 5.0 m/s [E] relative to the ground. a) What is Amanda's resultant velocity as witnessed by her friends on the shore? V A/W = 2.0 m/s [N] V W/G = 5.0 m/s [E] V A/G =? note VR = V A/G sketch V A/G = V A/W + V W/G V W/G V A/G VA/W VA/W W N θ V A/G OR W N α V W/G V A/G = V W/G + V A/W because order of vectors does not matter when adding... vector magnitude VA/G = VA/W 2 + VW/G 2 V A/G = (2.0 m/s) 2 + (5.0 m/s) 2 V A/G = (4.0 m 2 /s 2 ) + ( 25 m 2 /s 2 ) V A/G = 29.0 m 2 /s 2 V A/G = m/s vector direction tan θ = V W/G V A/W θ = tan m/s 2.0 m/ s θ = o θ = 68.2 o Final answer is V A/G = 5.4 m/s [N 68.2 o E] I could have worked with the second triangle but would have solved for α instead of θ. b) How much time does it take Amanda to swim across the river (assuming velocities remain constant)? Δdy = 750 m [N] Vy = Δdy Δt Δt= Δdy Vy Vy = V A/W = 2.0 m/s [N] Δt =? Δt = 750 m 2.0 m/s = 375 s = 3.8 x 10 2 s c) How far downstream would Amanda land? Δdx =? Vx = V W/G = 1.0 m/s [E] Δt = 3.8 x 10 2 m/s Vx = Δdx Δt Δdx = Vx Δt The time moving east equals the time moving north. Δdx = (5.0 m/s) (375 m/s) = 1875 m = 1.9 x 10 3 m [E]

9 If you read over example 1 and example 2, did you notice the following: 1. Amanda's resultant velocity increased in magnitude, 2. Amanda's resultant velocity is becoming more horizontal or eastward as the vector is moving towards the east axis and away from the north axis, 3. the time to cross was the same, and 4. Amanda's displacement downstream increases as the current speed increases.

10 Were you surprised that the time to cross was the same? This is an important point to understand. The water's movement itself, the current, is responsible for her eastward displacement. Amanda's speed relative to the water is the speed that will determine the time to cross from the south shore to the north shore. This means, if there was zero current, her resultant velocity would be 2.0 m/s [N]. It would take 375 s (ignoring significant figures here) to cross the 750 m divide. If the current is suddenly turned on, she is carried more to the east but it is still the northerly velocity (her velocity relative to the water) that explains her movement from the south to the north shore. Her resultant velocity must have a magnitude larger than that of her speed relative to the water because she travels a longer distance than the actual width of the river.

11 Perpendicular vector act independently of each other. For example, the horizontal displacement is determined by the horizontal velocity only. The vertical displacement is determined by the vertical velocity only. The horizontal acceleration is determined by the horizontal net force. The vertical acceleration is determined by the vertical net force. An animation that allows you to calculate the time to cross as you change speeds and directions. Navigating exercise: a fun animation offering different destinations.

12 Could you have used her resultant velocity to determine the time to cross? Yes, you can use the resultant velocity to determine the time to cross. But, it will add a few more steps to the process. V W/G Here is the diagram from example 1. V A/W V A/G W N θ Δdx If I wanted to use V A/G, I would first need to calculate her resultant displacement in the direction [N θ E]. Δd y W N θ Δd R I would use the angle I calculated, θ = 26.6 o, and the adjacent side, Δdy to calculate the Δd R. cos θ = Δdy ΔdR Δd R = Δdy = 750 m = m cos θ cos o Then, use the resultant displacement and the resultant velocity to calculate the time to cross. Again, these two vectors have the same direction! V R = Δd R Δt Δt= Δd R V R Δt = m m/s = s = 3.8 x 10 2 s

13 If you understood the last few pages, then try the next one.

14 Ex. 3 Suppose the friends see Amanda's direction of travel is due north. Amanda is swimming 2.0 m/s relative to the water. The river's current is flowing 1.0 m/s [E] relative to the ground. a) What is Amanda's heading (the angle at which she keeps her body) as crosses the river? b) How much time does it take Amanda to swim across the river (assuming velocities remain constant)?

15 Answers for example #3. a) 1.7 m/s [N 30 o W] or 120 o. If you didn't get the correct answer, I would bet you drew the wrong triangle. Go back and read the problem again very carefully. b) 4.3 x 10 2 s

16 Need a hint?

17 current's velocity resultant velocity velocity relative to water W E S Now, proceed again.

18 p. 110 Homework:

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