Question

a person walk at first constant speed 5 m/s then back alone the straight line from B to A at a constant speed 3 m/s a) what is his average speed over the entire trip b) what is his average velocity over the entire trip​

Answer 1

Answer:

3.75 m/s

Explanation:

Comment

Let's deal with the second question first. The average velocity is 0 because the displacement (distance between the starting point and ending point) is 0.

The first question is a little harder. You don't seem to have enough information. When that happens, you can just make things up. Now there's an interesting solution. You could do it with algebra, but it is easier to see with numbers.

Givens

• d = 150 m. between A and B
• r1 = rate from A to B = 5 m/s
• r2 = rate from B to A = 3 m/s
• time A to B = t1
• time B to A = t2

Formulas

t1 = d / r1

t2 = d/r2

average rate = total distance / total time

Solution

t1 = 150 m / 5 m/s = 30 seconds

t2 = 150 m / 3 m/s = 50 seconds

Total distance = 150 m + 150 m = 300 m

Total time = 30 seconds + 50 seconds = 80 seconds

Average speed = 300 m / 80 s

Answer

Average speed = 3.75 m/s