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paulb203

Member

Registered: 2023-02-24

Posts: 136

Calculating average velocity for an object in freefall

When an object in freefall, ignoring air resistance, accelerates from 0m/s to 1 m/s in its 1st second we are told that its average velocity is 5 m/s, and that it will have covered a distance of 5m. How is this calculated?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,199

Re: Calculating average velocity for an object in freefall

hi paulb203

There are a set of equations for distance travelled, initial and final velocity, time taken; all assuming the acceleration is constant.

I use these letters:

distance s ;   initial velocity u ; final velocity v ; time taken t ; acceleration a.

The formulas are:

s = ut + 0.5 a t^2  (i)

v = u  + at              (ii)

v^2 = u^2 + 2as     (iii)

average velocity = (u + v)/2    (iiii)

You are told u = 0 ; v = 1 ; t = 1 ; s = 5

accelerates from 0m/s to 1 m/s in its 1st second we are told that its average velocity is 5 m/s, and that it will have covered a distance of 5m.

Using (iii) 1 = 0 + 2a times 5  so a = 1/10

but using (ii) 1 = 0 + a times 1 so a = 1

Something funny is going on here.

You say freefall so I'll assume an object falling to Earth. The acceleration due to gravity is 9.81 m/s/s which is approxiamtely 10 so let's use that figure.

If the object starts with u = 0 then after one second v = 0 + a times 1 = 10.

Now we have average velocity = (0 + 10)/2 = 5 which makes more sense.

And s = o.t + 0.5 a t^2 = 5 which again makes sense.

So maybe that v = 1 m/s is a typo.

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

paulb203

Member

Registered: 2023-02-24

Posts: 136

Re: Calculating average velocity for an object in freefall

Another thorough, clear answer. Thanks a lot, Bob.

First off, yes the 1m/s final v was a typo, and yes, it should have been 5m/s

I’ve had a go at plugging the values in to your 3 formulae;

s = ut + 0.5 at^2
5m = 0m/s * 1s + 0.5 * 10m/s/s * 1^2
5m = 0.5 * 10
______________________________________________________
v = u + at
10m/s = 0m/s + 10m/s/s * 1
10m/s = 10m/s/s
_____________________________________________________
v^2 = u^2 + 2as
10m/s^2 = 0m/s^2 + 2 * 10m/s/s * 5m
10m/s^2 = 2 * 10 * 5
100 = 20 * 5
____________________________________________________
Was there a typo in your final part?
“And s = o.t + 0.5 a t^2 = 5 which again makes sense.”

___________________________________________________

“Now we have average velocity = (0 + 10)/2 = 5”

Ah, so that’s how the average velocity is worked out...

So for the next couple of seconds would it be;

1-2 seconds; From 10m/s to 20 m/s; average v =
(10+20)/2 = 15m/s

2-3 seconds; From 20m/s t0 30 m/s; average v =

(20+30)/2 = 25m/s

Then 70/2 = 35; 90/2 = 45; etc, etc?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,199

Re: Calculating average velocity for an object in freefall

hi paulb203

Those additional calculations look good to me.

And s = o.t + 0.5 a t^2 = 5 which again makes sense.

Maybe because I didn't put it in maths formatting or maybe because I typed a little o rather than a zero.

u=0, t=1, a=10 so

and that's what we had elsewhere for s.

Those formulas come from Newton's laws.

If you draw a velocity time graph and a = constant then the velocity will go up in a straight line with gradient 'a'.

so we have ds/dt = v = u + a.t

If you integrate this wrt t you get

s = u.t + half a t^2 + constant.  easy to make the constant zero by choosing to measure s from when t=0

Because a = constant the average v will just be (start v + end v)/2

Finally t = (v-u)/a

Eliminate t in the 's' formula:

s = u(v-u)/a + 0.5 a (v-u)^2/(a^2) => as = u(v-u) + 0.5 (v-u)^2

2as = 2uv - 2u^2 + v^2 + u^2 -2uv = v^2 - u^2 and so the v^2 formula drops out from 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

paulb203

Member

Registered: 2023-02-24

Posts: 136

Re: Calculating average velocity for an object in freefall

Thanks again, Bob.