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I'm working with rotational motion, and one of the mistakes that I made in solving for distance in terms of time was substituting angular acceleration for (d/Rt^2) where d is distance and R is the radius of the wheel. I proceeded to solving for d.

Apparently the correct method is to apply rotational kinematics, which I don't believe I understand conceptually.

Going back to translational kinematics, If I want to solve the distance traveled in time t with an acceleration of 3, I would use d = 1/2(3)t^2, assuming initial velocity and starting distance was 0. Why does it not simply work to use the literal definition of acceleration (d/t^2) and say that 3 = (d/t^2) to solve for d? You get two different answers, and I know that the former is correct, however, I do not understand why. I feel as though an understanding here will carry over to the problem I am trying to solve that involves rotational kinematics.

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    $\begingroup$ Definition of acceleration is not $a=d/t^2 $. Where did you see this? $\endgroup$
    – nasu
    Commented Oct 20, 2019 at 0:21
  • $\begingroup$ To me, it looks like your core question is: why is the formula for displacement $s=\frac12 at^2$, where the displacement $s$ and velocity at time $t=0$ are both zero, and the acceleration $a$ is constant. Did your teacher explain that formula, or was it just stated without proof, as something you just have to memorize? $\endgroup$
    – PM 2Ring
    Commented Oct 20, 2019 at 0:42

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There are two points here, I believe: one is that acceleration is something that is defined for very short times. It tells us the rate of change in velocity, so we can think of it as "how much is $v(t+\delta t)$ different than $v(t)$ for very small $\delta t$?"

When deriving the formula you are referring to, of $x=\frac{1}{2}at^2$, an assumption was made that the acceleration was constant all the time, and therefore we can simply integrate to get the distance traveled over straight line $\Delta x = \int_0^{t} dt' v(t')$ and then use $v(t) = \int_0^{t}dt' a = at$ to get the result you mentioned. This is not true in general, when acceleration can change or in higher dimensions.

A second point, and possibly unrelated, is the question of the rotational motion. I'm not sure I fully understand the problem you presented but it seems that you were asked to calculate the distance that a rotating disk will travel if it had a constant angular acceleration? In this case the question is how to translate the angular acceleration - which is how faster the disk rotates around itself - to the distance the disk traveled. Note that angular acceleration is in units of radians per time squared, while linear acceleration is in units of distance per time squared. To get one from the other we have to use the radius of the disk. A large disk rotating at the same rate will cover more distance than a small disk (this is the principle on which gears in bicycles work). A disk of radius $R$ which rotates at angular velocity of $\omega$ will travel at a linear velocity of $\omega R$. Now we can do the same and derive from the angular acceleration what would be the linear acceleration, and if it is indeed constant then we can use the formula you were given. (Assuming that I understood the problem that you were given)

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  • $\begingroup$ Note that the OP uses $d$ as a variable name, so it's very likely that they haven't learned any calculus yet. $\endgroup$
    – PM 2Ring
    Commented Oct 20, 2019 at 9:22

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