A person walks at constant speed of 5m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 3 m/s . (a)her average speed over the entire trip ? (b) her average velocity over the entire trip ?
Note : average speed = distance travelled / time
2006-10-19 18:45:56 · 10 answers · asked by mr. h 1 in Science & Mathematics ➔ Physics
Let t1 be the time taken to go from A to B.
Distance traveled in t1 is 5t1.
Let t2 be the time taken to go from B to A.
Distance traveled in t2 is 3t2.
But 5t1 = 3t2. Since the distances are the same.
Total distance traveled is 2 x 5 t1 OR 2 x 3 t2.
Total time is t1 + t2 = t1 + 5/3 t1 = 8/3 t1.
Or 3/5 t2 + t2 = 8/5 t2.
Average speed is (total distance / time) = 6t2 / (8/5 t2)
= 3.75 m/s.
Or 10 t1/ (8/3 t1)= 3.75 m/s.
Since the starting and finishing points are the same the total displacement is zero and therefore the average velocity is zero.
2006-10-20 01:20:13 · answer #1 · answered by Pearlsawme 7 · 0⤊ 0⤋
Speed and velocity are not the same. Velocity has a direction, speed is a number. Speed is the "magnitude" of velocity. The average speed is how far she walked divided by how long it took. If the distance is AB meters, it took her AB/5+AB/3 seconds to complete the trip. The distance is 2AB so her average rate was distance/time = 15/4 m/s. (this makes sense since she spends more time walking at 3m/s than she spends walking at 5 m/s).
The average velocity is the "displacement" divided by time. Displacement is how far away you are from when you started, which is zero in this case. The average velocity is zero. This makes sense since her velocity during the forward trip is in opposite direction to her velocity in the return trip.
Update_
I thought it was obvious that i was refering to opposite direction, since magnitudes don't have opposites. Regardless, I've clarified the point.
2006-10-20 02:40:48 · answer #2 · answered by lorentztrans 2 · 1⤊ 0⤋
Thhe average speed should be (5+3)/2 as no information about distance and time is provided. About average velcity as the person is travellin in straight lin espped is same as the velocity. so from A to B the velocity will be +5m/s but from B to A ti will be -3m/s. So now you have enough information to solve the sum. But i doubt the the person being a girl for it's crazy.
2006-10-20 02:22:03 · answer #3 · answered by dude 1 · 0⤊ 0⤋
speed = distance /time
(a) time (1)=AB/5 time(2)=BA/3
total time=(AB*3+BA*5)/15 =8*AB/15 (AB=BA as a scalar quantity)
total speed=2AB/(8*AB/15)=3.75m/s
This is the old you cant take an average of an average trick.
(b)velocity=speed in a certain direction, eg in the direction AB.
you can add these speeds as vectors 5m/s-3m/s=2m/s
(later)
Lorentztr says her velocity in the foward trip is the same magnitude as in the return trip. Only her DISPLACEMENT is the same magnitude in either direction.
2006-10-20 02:30:07 · answer #4 · answered by sydney m 2 · 0⤊ 0⤋
constant speed of 5m/s uh if that speed is constant isn't that the average speed? and average velocity for that matter??
2006-10-20 01:56:23 · answer #5 · answered by Den P 3 · 0⤊ 0⤋
i donot abt average speed bcz neither i know distance nor time but A TO B its speed will 5M/S and from b to a its speed will be 3m/S BUT ITS VELOCITY WILL BE ZERO BCZ BODY HAS NOT TRAVELLED IN ONE STRAIGHT DIRECTION OR SPECIFIED ONE
2006-10-20 02:24:01 · answer #6 · answered by ghulamalimurtaza 3 · 0⤊ 0⤋
speed and velocity are same
the average speed is 4m/s so average velocity is also 4m/s
average speed =
(5m/s+3m/s)/2 = 4m/s
2006-10-20 02:22:11 · answer #7 · answered by Navdeep B 3 · 0⤊ 0⤋
I'm totally thrown off by the example being a girl. It should be a guy.
2006-10-20 01:48:30 · answer #8 · answered by Anonymous · 0⤊ 1⤋
a) 3 m/s (scalar)
b) 0 m/s (vector)
Th
2006-10-20 01:51:03 · answer #9 · answered by Thermo 6 · 0⤊ 0⤋
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2006-10-20 01:51:57 · answer #10 · answered by Anonymous · 0⤊ 0⤋