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**paulb203****Member**- Registered: 2023-02-24
- Posts: 25

On this website, and many others, the velocity of a freefalling object in a vacuum is shown as follows.

After 1,2,3 and 4 seconds respectively;

9.8m/s; 19.6m/s; 29.4m/s; 39.2m/s

I worked out the distance travelled by a freefalling object in a vacuum using d = at^2/2, or, d = 0.5gt^2 and got, for after 1,2,3 and 4 seconds;

4.9m; 19.6m; 44.1m; 78.4m

But when I tried to double check the above using the formula for velocity, v= ΔX/t, I got, for after 1,2,3 and 4 seconds;

4.9m/s; 9.8m/s; 14.7m/s; 19.6m/s

Where have I gone wrong?

I’m always trying to bear in mind average velocity versus instantaneous velocity and I’m aware that there are several formulas for velocity but I thought those were mostly rearrangements of each other.

I see though that v=at works for the velocities at the top of this page.

How do v= Δx/t and v=at relate to each other?

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,718

hi paulb203

Those calculations are all correct. The velocity at each instant will not be the same as the average velocity over a period as the object is accelerating.

Use v = u + at for the instantaeous velocity at a moment and average v = d/t for the average velocity (or speed is a better term for this)

Bob

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

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**paulb203****Member**- Registered: 2023-02-24
- Posts: 25

Thanks, Bob. I wish the website where I got v=at had put a subscript after v indicating that this equation only works for instantaneous velocity

The equation v=delta x/t usually has a bar line over the v to indicate average v, yes?

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,718

I've not noticed that as a usual practice. To make it clear I prefer to say 'average speed'

Velocity is a vector measure; direction is part of the measure along with magnitude. If an object is dropping vertically then the direction isn't changing but in other cases it might well be. If you computed the distance travelled divided by the time taken this would not take any account of this varying direction. It is for this reason that I would use speed, which is scalar, rather than velocity.

Bob

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

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