- #1

Ocata

- 198

- 5

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.

[itex]\int a (dt) = at +v_0 = v_f[/itex] ,

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

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

[itex]r(3) = \frac{1}{2}(1)3^2+0+0=4.5[/itex] 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?