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RELATIVE VELOCITY

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.

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RELATIVE VELOCITY

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

  2. Notation • vA = velocity of object A with respect to a stationary object. • vAG= velocity of object A with respect to ground (stationary object on the ground) • vA and VAG have the same meaning. • vAB=velocity of object A in the frame of reference (with respect to) object B • vAB=vA-vB(velocity of object A with respect to ground - velocity of object B with respect to ground.)

  3. Relative Velocity (One Dimension) A B vA=+5.0 m/s vB=0 m/s vAB=vA-vB =+5.0 m/s – 0 m/s = +5.0 m/s vBA=vB-vA = 0-(+5.0m/s) = -5.0 m/s vAB = -vBA(Always)

  4. 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 vB=+5.0 m/s vA=-10 m/s vAB=vA-vB = -10 m/s – (+5.0 m/s) = -15.0 m/s vBA = +15.0 m/s

  5. 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 vB=+5.0 m/s vA=+10 m/s vAB=vA-vB = +10 m/s – (+5.0 m/s) = +5.0 m/s vBA = -5.0 m/s

  6. Relative Velocity(One object moving with another object) vBC =velocity of ball with respect to car vCG=velocity of the car with respect to ground vBG=velocity of ball with respect to ground vCG=+10 m/s vBC=+20 m/s vBG=vBC+vCG vBG=+20m/s +(+10 m/s)= +30 m/s

  7. Check of Relative Velocities • vBG=vBC+vCG • vBG=(vB-vC)+(vC-vG)=vB-VG

  8. Relative Velocity(Two Dimensions) Situation 1: vA θA B A θB vB vABx=vAcosθA-vBcosθB vABy=vAsinθA-vBsinθB

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

  10. Relative Velocity in Two Dimensions Situation 3: vWG vWG vBW vBG=vR vBW θ vBG=vBW+vWG θ=sin-1(vWG/vBW) vR = resultant boat speed vR The angle needed to travel directly across the stream.

  11. The angle necessary to dock a specific distance downstream: vWG Situation4: x θ d vW vB α 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: vW,,vB, x, and d. 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. vBG=vB + vW = vR

  12. Situation 4 continued x z θ d vW z α vBG=vR vB φ vBG vR = resultant velocity from water and boat β Use law of sines. vW/vB=sin φ/sin z, solve for φ α = θ - φ

  13. Distance to dock upstream. x Situation 4a α d vW vBG vBG=vR φ θ vR = resultant velocity from water and boat vB

  14. Distance downstream docked based on specific boat direction: x Situation 5 z θ d vWG z α vBG=vR φ vBG vB vR = resultant velocity from water and boat Known: vWG, vBW, d, and α. How far will the boat aimed at an angle, α, dock downstream?

  15. Situation 5 continued x z θ d vBG=vR vBx vW vR = resultant velocity from water and boat z α = vBx φ vBG vB vRx=vW+vB(sin α) vRy = vB(cos α) θ=tan-1(vRx/vRy) tan θ=(x/d)  solve for x.

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