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2D Relative Motion

2D Relative Motion. River and Fight Problems. River Problems. Terms : River (current) velocity V c – speed and direction river is flowing Boat velocity V b – speed and direction boat is pointed

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2D Relative Motion

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  1. 2D Relative Motion River and Fight Problems

  2. River Problems Terms : • River (current) velocity Vc– speed and direction river is flowing • Boat velocity Vb– speed and direction boat is pointed • Ground (resultant) Velocity Vr – speed and direction boat travel in relation to an observer on the bank.

  3. Solving River Problems • Set up vector diagram of velocities, making sure boat is pointing in correct direction • Place info given • Solve for missing velocity (usually ground, but not always!) • Remember that time is the same for all components of the problem (usually give as time taken to cross) • Displacement components will have same direction as velocity components • Solve for Displacement or time as required • E.g. A boat with a velocity of 12.0 km/h points [N] across a river with a current of 5.0 km/h E. • What will its resultant velocity be? • Determine how long it will take to cross the river if the river is 350 m wide • How far down river will it land? • What direction will it have to move straight across the river? • How long will it take to cross if it moves straight across?

  4. River Problem Example : A Resultant Velocity • Use Pythagorean Theorem to find magnitude: vr= vr= 13 km/h • Use Trig to find direction: Tan  = 12/5   = Tan-1 (12/5)  = 67  N of E • The resultant Velocity is 13 km/h 67  N of E Vr = ? Vb = 12 km/h [N]  Vc = 5 km/h [E]

  5. River Problem Example : B • We want the time to cross the river which is 350m straight across so we use the velocity in the same direction, vb! • v = d/t  t = d/v • t = 350m/3.33333m/s * • t = 105s = 1.1 x 101 s = 1.8 min * note converted vb= 12 km/h to 3.33333333m/s 350 m vb

  6. River Problem Example : c • We want the distance down river, so now that we know the time, we can use the current velocity vcto find this! • v = d/t  d = vct • d = 1.388888m/s* (105s) • d = 146 m = 1.5 x 101 m * note converted vc= 5 km/h to 1.3888888889m/s Distance down river 350 m

  7. River Problem Example : D Resultant Velocity - now is not the hypotenuse, vbis! • Use Pythagorean Theorem to find magnitude: vr= vr= 10.9 km/h = 11 km/h • Use Trig to find direction: Cos  = 5/12   = Cos-1 (5/12)  = 65  N of W • The resultant Velocity is 11 km/h due north! • The boat is pointed 65  N of W!! Vb = 12 km/h [N] Vr = ?  Vc = 5 km/h [E]

  8. River Problem Example E: • We want the time to cross the river which is 350m straight across so we use the velocity in the same direction, vrin this case! • v = d/t  t = d/v • t = 350m/3.0302m/s * • t = 115.5s = 1.2 x 101 s = 1.9 min * note converted vr= 10.9 km/h to 3.0302m/s vr 350 m

  9. Flight Problems Are very similar, except in air instead of water! Term: • Wind velocity Vw– speed and direction wind is blowing to (a west wind blows from the west – so its direction is East !) • Air velocity Va– speed with respect to air and direction aircraft is pointed • Ground (resultant) Velocity Vr – speed and direction boat travel in relation to an observer on the ground. • E.g. a bird travelling at 4.2 m/s [S] while an east wind blows at 1.5m/s vr va vw= 1.5m/s [W] va= 4.2m/s [S] vw

  10. Flight Problems: Example We could calculate all similar components for this problem, but lets change it up a bit. • Calculate the air speed and direction the bird would have to maintain to travel due south at 4.0 m/s in an east wind of 1.5m/s. • How long would it take to fly 1.o km in this case? • Note: we are given the resultant now and the bird must fly into the wind (the air velocity is the hypotenuse)!  va= ? vr= 4.0m/s vw= 1.5m/s vr= vr= 4.2720 = 4.3 m/s Use Trig to find direction: Tan  = 1.5/4.0   = Tan-1 (1.5/4.0)  = 21  E of S The air velocity of the bird must be is 4.3 m/s 21  E of S

  11. Flight Problems: Example (cont.) • How long would it take to fly 1.o km in this case? • Since the bird is actually travelling south at 4.0m/s, that is the velocity we use!  va= ? vr= 4.0m/s vw= 1.5m/s v = d/t  t = d/v t = 1000m/4.0m/s * t = 250s = 2.5 x 101 s = 4.1 min (pretty quick bird!) * note converted 1.0 km to 1000m

  12. A note about non-perpendicular vectors • The examples used here have always been perpendicular….but that will not necessarily always happen e.g. a plane flies at 450km/h @ 45 S of E in a wind of 100km/h coming from 60 N of E. • To find the resultant velocity you will need to use vector components! vrx  vr vry vay va vw vwy vax vwx

  13. Assignment • Do pg 95 and 97 practice problems

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