RELATIVE VELOCITY The velocity of an object with respect to another object.

Notation v A = velocity of object A with respect to a stationary object. v AG = velocity of object A with respect to ground (stationary object on the ground) v A and V AG have the same meaning. v AB =velocity of object A in the frame of reference (with respect to) object B v AB =v A -v B (velocity of object A with respect to ground - velocity of object B with respect to ground.)

Relative Velocity (One Dimension) A B v A =+5.0 m/sv B =0 m/s v AB =v A -v B =+5.0 m/s – 0 m/s = +5.0 m/s v BA =v B -v A = 0-(+5.0m/s) = -5.0 m/s v AB = -v BA (Always)

Relative Velocity (Two Objects Moving in Opposite Directions) The relative velocity of two objects moving in opposite directions is the sum of the two speeds with the appropriate frame of reference direction. A B v A =-10 m/s v B =+5.0 m/s v AB =v A -v B = -10 m/s – (+5.0 m/s) = m/s v BA = m/s

Relative Velocity (Two Objects Moving in the Same Direction) The relative velocities of two objects moving in the same direction is the difference of the two speeds and with the appropriate frame of reference direction. A B v A =+10 m/s v B =+5.0 m/s v AB =v A -v B = +10 m/s – (+5.0 m/s) = +5.0 m/s v BA = -5.0 m/s

Relative Velocity (One object moving with another object) v BC =+20 m/s v CG =+10 m/s v BG =v BC +v CG v BC =velocity of ball with respect to car v CG =velocity of the car with respect to ground v BG =velocity of ball with respect to ground v BG =+20m/s +(+10 m/s)= +30 m/s

Check of Relative Velocities v BG =v BC +v CG v BG =(v B -v C )+(v C -v G )=v B -V G

Relative Velocity (Two Dimensions) A B θAθA θBθB v ABx =v A cosθ A -v B cosθ B v ABy =v A sinθ A -v B sinθ B vAvA vBvB Situation 1:

` Relative Velocity in Two Dimensions v WG v BW v WG v BG v BG =v BW +v WG θ θ=tan -1 (v WG /v BW ) What is the velocity of the boat in the earth’s frame of reference? Situation 2: v WG = velocity of water with respect to groundv BW = velocity of boat with respect to water v BG = velocity of boat with respect to the ground.

Relative Velocity in Two Dimensions v WG v BW v WG v BG =v R v BG =v BW +v WG θ θ=sin -1 (v WG /v BW ) The angle needed to travel directly across the stream. Situation 3: vRvR v R = resultant boat speed

v WG vBvB v BG =v B + v W = v R θ=tan -1 (x/d) where θ=α+φ, which is the resultant angle of the boat. α= boat angle, φ=angle between boat’s aimed direction (boat angle) and actual direction. α The angle necessary to dock a specific distance downstream: x d φ What angle, α, must the boat be directed to dock a distance, x, downstream while crossing a river that is a distance, d, wide? Known: v W,,v B, x, and d. θ vWvW Situation4:

vWvW vBvB α x d φ θ β v BG v BG =v R z z Use law of sines. v W /v B =sin φ/sin z, solve for φ α = θ - φ v R = resultant velocity from water and boat Situation 4 continued

vWvW vBvB α x d φ θ v BG v BG =v R v R = resultant velocity from water and boat Distance to dock upstream. Situation 4a

v WG vBvB α x d φ θ v BG v BG =v R z z v R = resultant velocity from water and boat Distance downstream docked based on specific boat direction: Known: v WG, v BW, d, and α. How far will the boat aimed at an angle, α, dock downstream? Situation 5

vWvW vBvB α x d φ θ v BG v BG =v R z z v R = resultant velocity from water and boat v Rx =v W +v B (sin α)v Ry = v B (cos α) = v Bx tan θ=(x/d)  solve for x.θ=tan -1 (v Rx /v Ry ) Situation 5 continued v Bx