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 2.70 m/s.
(a) What is her average speed over the entire trip?
(a) What is her average speed over the entire trip?
Answers
drwls
It is not the average of 4.9 and 2.7, because she spends more time walking at the slower speed.
Let the total distance between A and B be D
t1 = D/4.9 t2 = D/2.7
Average speed = 2D/*t1 + t2)
= 2D/[(D/4.9) + (D/2.7)]
= 2/[(1/4.9) + 1/(2.7)] = 3.38 m/s
Let the total distance between A and B be D
t1 = D/4.9 t2 = D/2.7
Average speed = 2D/*t1 + t2)
= 2D/[(D/4.9) + (D/2.7)]
= 2/[(1/4.9) + 1/(2.7)] = 3.38 m/s
Beti
Ur great but the solution is like .....