# Vt=d doesn't seem to hold true w/ constant acceleration

1. Jun 17, 2015

### Ocata

Hello,

Lets say I start off with constant acceleration of 1 m/s^2, and I want to evaluate it at 3 seconds.

a = 1.
a(3) = 1

then, since at=v,

v(t) = at = 1(3) = 3m/s then I can work backward v/t = a = 3/3 = 1

then, since vt = d

d(t) = vt = 3(3) = 9 then I can work backward d/t= v = 9m/3s = 3m/s

This way, the units make sense acceleration is 1 m/s/s

Take 9 meters. Divide it by 3 seconds. Then divide it again by 3 seconds. And voila, you arrive back at 1m/s/s

_______

However, if I integrate acceleration to get the kinematic equations, it all seems to fall apart and I don't know why.

$\int a (dt) = at +v_0 = v_f$ ,

so evaluating time = 3 with acceleration 1m/s^2,

v(3) = 1(3)+0 = 3m/s then I can work backward v/t = a = 3/3 = 1 => all good still

then

$\int v (dt) = \frac{1}{2}at^2 +v_0(t)+r_0= r_f$,

$r(3) = \frac{1}{2}(1)3^2+0+0=4.5$ so if I work backward this time d/t=v = 4.5/3= 1.5 m/s

But my velocity is 3m/s, not 1.5m/s

So my conclusion is, Distance/Time = Velocity will only hold true if there is 0 acceleration. Is this correct?

2. Jun 17, 2015

### Staff: Mentor

Or if v is average velocity.

Edit: which is what you get from integrating.

Last edited: Jun 17, 2015
3. Jun 17, 2015

### ZapperZ

Staff Emeritus
Nope. This is wrong. This is only true if v is a constant. Since your acceleration is not zero, v is NOT a constant, so this is incorrect.

Think about it, If you have a constant v, you use this equation. You have a non-constant v, you ALSO used this equation. Something's got to give because it can't both be correct.

Zz.

4. Jun 17, 2015

### Ocata

Hi ZapperZ,

Thank you for responding.

May I ask, which part is wrong exactly? I am betting that just the first part is wrong and that the second part (which contradicts the first part) is correct, since my final conclusion is:

I feel that by the end of the work, you are in agreement with my final conclusion. But maybe that you are correcting another point I have missed.

When solving for velocity using each formula, at = v = 3m/s and at + v_0 = 3m/s,

both arrive at the correct answer, but this is only when assuming initial velocity is 0.

But it breaks down when solving for distance, regardless of whether I have initial velocity or initial position.

Is saying "constant velocity" the same thing as saying "0 acceleration."?

5. Jun 17, 2015

### ZapperZ

Staff Emeritus
Look at the last line I quoted in your post. The use of d=vt is invalid for varying v.

Zz.

6. Jun 19, 2015

### Ocata

Thank you. Much appreciated.

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