A roulette ball spins around a rim.

What is the time ($t$) and distance ($d$) in which the velocity of the ball $= v$ (whatever number)?

The Deceleration of the ball from $4$ spins was plotted on a scatter and was best described with the following $3$rd order polynomial:

$$A(t)= 0.0396t^3 - 1.1035t^2 + 10.25t -33.898 \quad\text{with}~ R^2 = 84.6%$$

I have been trying to find Time and distance from velocity and acceleration using:

\begin{align*}t &= \frac{\text{Final Velocity} - \text{Initial Velocity}}{\text{Acceleration}} d \\ &= \text{Initial Velocity} \times \text{Time} + \frac12 \text{Acceleration} \times \text{Time}^2 \end{align*}

Unfortunately this equation assumes a constant acceleration and so gives me nonsense since the acceleration of the ball is not constant.

I don't know if this will be useful but the relationship I see is: the higher the ball velocity, the higher the rate of deceleration. So I plotted Acceleration as a function of Velocity:

$$A(V) = -0.00008v^3 + 0.0125v^2 - 0.7688v + 14.983 R^2 = 0.88$$

Thank you I've researched this for 3 damn hours.. if anyone can put me in the right direction it would be appreciated, thanks

  • $\begingroup$ I hope you have had a course in calculus because, without it, you cannot solve this problem. $\endgroup$ – ja72 Dec 2 '18 at 5:45
  • $\begingroup$ Your formula is only applicable for constant acceleration. $\endgroup$ – amd Dec 2 '18 at 5:55
  • $\begingroup$ I did calculus a few years ago, im a bit rusty. especially since my applicaiton of calculus was for econ and econometrics, not physical problems... and also mainly only first or second derivatives. $\endgroup$ – Tom Edouard Pelletier Dec 2 '18 at 17:21


For continuously varying acceleration, use the relation $a(t)=\frac{d[v(t)]}{dt}\implies v(t)=\int_{t_1}^{t} a(t)\ dt+v(t_1)$

Once you obtain velocity as a function of time $t$, use the relation $v(t)=\frac{d[s(t)]}{dt}\implies s(t)=\int_{t_1}^{t} v(t)\ dt+s(t_1) $

As far as finding the time for a particular velocity or acceleration is concerned, you know the expressions for $v,a$ as functions of time. You just need to plug-in the values and find the roots of the polynomial in $t$.


In your case acceleration as a function of time only, $a(t)$. So direct integration applies here:

$$ \begin{aligned} & \text{speed} & & \text{distance} \\ v(t) & = v_0 + \int \limits_{t_0}^t a(t)\,{\rm d}t & x(t) & = x_0 + \int \limits_{t_0}^t v(t)\,{\rm d}t \end{aligned} $$


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