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A slower moving car is traveling behind a faster moving bus. The velocities of the two vehicles are: vCG = velocity of the Car relative to the Ground = +12 m/s vBG = velocity of the Bus relative to the Ground = +16 m/s A passenger on the bus gets up and walks toward the front of the bus with a velocity of vPB, where vPB = velocity of the Passenger relative to the Bus = +2 m/s What is vPC, the velocity of the Passenger relative to the Car?

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obasola1

Answer:

6m/s

Explanation:

A slower moving car is traveling behind a faster moving bus. The velocities of the two vehicles are: vCG = velocity of the Car relative to the Ground = +12 m/s vBG = velocity of the Bus relative to the Ground = +16 m/s A passenger on the bus gets up and walks toward the front of the bus with a velocity of vPB, where vPB = velocity of the Passenger relative to the Bus = +2 m/s What is vPC, the velocity of the Passenger relative to the Car?

When you are is in a car moving with 12m/s speed and then there is a bus moving with 16m/s in front of you. Of course, the bus is 4m/s faster than you. if a passenger in the bus stands and runs towards the front of the bus with a speed of 2m/s , he is 2m/s ahead +the 4m/s the bus was already ahead of you, mathematically

Vpg=speed of the passenger relative to the ground

Vpb=sped of the passenger relative to the bus

Vbc=speed of the bus relative to the car

Vbg=speed of the bus relative to the ground

Vpg=Vpb+Vbc

Vbc=Vbg+Vgc

so we can conclude that

Vpc=Ypb+Vbg+Vgc

Vpc=2+16-12

Ypc=6m/s

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