Help with a physics question about average velocity?

Question

I have a question about the process of this problem:

A skydiver with an unopened parachute falls 625 m in 15 s. Once the parachute is opened, they fall another 356 m in 142 s. What is the average velocity for the entire fall?

Do you just find the velocity for the individual sections, and then use the average velocity formula?

Wait, why 981 m/ 29 s? How do you get the 29 s?

Answer 1

Yeah, its sort of a trick question. Average velocity is just concerned with total distance traveled and total time traveled.
They actually fell 157 s not 29. So it's 981/157.

Answer 2

He fell 981m in 29s, so his average velocity is 981/29 ....

or 33.83 m/s.

Answer 3

It was Paris that discovered the definiton of average velocity.
So the same principle apply to parachuting.

Answer 4

So i visit only inform you the approach to fixing it : First you need to calculate the excellent displacement travelled, via multiplying the speed to time then you definately ought to sum up the excellent time (upload at the same time) finally, v = s/t = displacement / time So divide the excellent displacement you acquire with the excellent time you acquire it incredibly is customary speed P.S. do no longer overlook that speed is a vector, so be confident to grant the guidelines ! (East)

Answer 5

just to make sure you choose the right answer, Ash has the right idea here. Add the distances, and the times for each, and devide them.

Answer 6

Yep tomgreen got it.

Answer 7

yes