"A person first walks at a constant speed V1 along a straight line from A to B and then back along the line from B to A at a constant speed v2" What is her average speed over the entire trip?
I thought it was just (V1+V2)/2 but someone told me you can't do that in this case?
2007-08-28 13:48:35 · 2 answers · asked by music123 1 in Science & Mathematics ➔ Physics
It's 2/(1/V1 + 1/V2). Call the one-way distance x. Then the time taken to walk 2x is x/V1 + x/V2. So average speed is
2x/(x/V1 + x/V2). Divide numerator and denominator by x and you have 2/(1/V1 + 1/V2).
2007-08-28 13:58:23 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
If the question asks for the average speed like you say then you were right. However, if it asked for the velocity your not.
You see, velocity is speed in a direction. An example of..
speed- 55mph
velocity- 55mph east
In this particular case if it is asking for velocity the answer should be zero because the person returned to the same spot they started out from. On average they didn't travel in any direction, this is why velocity would be zero.
2007-08-28 21:12:34 · answer #2 · answered by collective 2 · 0⤊ 0⤋