Question

a person walks first at a constant speed of 4.90 m/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 2.85 m/s

1.what is her average speed over the trip?
2.what is her average velocity over the entire trip?

Answer 1

do you think 1 is 5.7?

Answer 2

o wait jk
its 7.75

Answer 3

If you add up her two speeds and divide by two you get her average speed of 3.875m/s.
Her velocity is 0 for the reason that once she comes back to point A she hasn't moved and therefore her velocity is 0 (velocity = displacement/time)

Answer 4

it says the average is wrong

Answer 5

velocity is right

Answer 6

I'm sorry; can't help you then. Got no idea what I'm doing wrong; goodluck.

Answer 7

k

Answer 8

r t = d, t=d/r, r = d/t

Let d be one m, a convenient number since the distance is not defined in the problem statement.

The following expression divides the total distance walked by the total time consumed for the round trip walking.\[\text{AverageSpeed}=\frac{2}{\frac{1}{4.90}+\frac{1}{2.85}}=3.60\text{ }m/s \]

Answer 9

thats wrong too lol

Answer 10

@ riptidej

Would you be so kind as to provide the "right" answers here and justify them.
Thanking you in advance.

Answer 11

it doesnt tell me what is right or wrong lol

Answer 12

Then how do you know if they are right or wrong?

Answer 13

riptidej, riptidej,
Don't bailout on us now !