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paulb203

Member

Registered: 2023-02-24

Posts: 375

Average velocity

First second of freefall

According to physicsclassroom.com;

Using 10m/s/s for acceleration due to gravity near the surface of Earth;

Time interval=1s
Change in v=0-10m/s
Avg.v=5m/s
D=5m

I’m thinking the average v is arrived at using the mean of velocities;
(0+10)/2=5

Is that correct?

But to check, I’ve divided the first second into tenths

After 1/10sec, v=1m/s
After 2/10, v=2m/s
Etc, up to,
After 10/10, v=10m/s

But the mean of those is 5.5, I think;
1+2+3 etc, up to, +10 = 55
55 tenths = 5.5
Where have I gone wrong?

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Prioritise. Persevere. No pain, no gain.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,701

Re: Average velocity

If you draw a velocity-time graph it goes from (0,0) to (1,10)

The distance travelled is the area under the graph; 0.5 x 1 x 10 = 5m.

The average velocity is total distance travelled / time taken = 5/1 = 5.

So that's the correct answer.

From 0 sec to 1/10 sec the area under is

0.5 x 1/10 x 1  = 0.05 m = distance travelled

so the velocity over that interval is 0.05 / 0.1 = 0.5 m

You are using 1m ... that value is too big.  The correct value for each interval is the half way one not the endpoint one.

This happens for every interval from 0.1s to 1s.

So the correct average by this method is

(0.5 + 1.5 + 2.5 + 3.5 + 4.5 + 5.5 + 6.5 + 7.5 + 8.5 + 9.5)/10 =(5 + 45)/10 = 5 m/s

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

paulb203

Member

Registered: 2023-02-24

Posts: 375

Re: Average velocity

Thanks, Bob.

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Prioritise. Persevere. No pain, no gain.