Symbolic representation for time it takes for a variable to approach a specific value from an initial value

Question

For example, I came up with the following problem "If we have distance $d$ with an initial value of $2 \space feet$, with a secondary value of $10 \space feet$, and it takes us a time $t$ of $64 \space seconds$, what is the acceleration?"

I originally had it written down as this:

$$d = 2 \rightarrow 10$$ $$t = 64$$ $$d_{net} = 8$$ $$\frac{at^2}2 = d_{net}$$ $$\frac{a\cdot 64^2}2 = 8$$ $$\frac{4096a}2 = 8$$ $$2048a = 8$$ $$a = \frac{8}{2048} = \frac{1}{256} = \space .00390625$$

My question is for the $d = 2 \rightarrow 10$ and the $t = 64$ parts, how would one clearly symbolically represent the fact that we are talking about the time it takes $d$ to go from $2$ to $10$?

My guess would be something like:

$$d = 2$$

$$t(d \rightarrow 10) = 64 \space seconds$$

The only thing is that $t$ in this case is not a function, but a value we know. We are saying that the time it takes our distance to increase from $2$ to $10$ is $64$ seconds (and we use this in proceeding calculations). Sometimes, though, $t$ may be a function to determine our time, so what would be different in that case?

Answer 1

If you are dealing with problems of constant acceleration $a$, the main equations are: \begin{align} x(t) &= x(0) + v(0) t + \frac{a}{2}t^2 \quad, \forall t \geq 0 \\ v(t) &= v(0) + at \quad, \forall t \geq 0 \\ a(t) &= a \quad, \forall t \geq 0 \end{align} where $x(t)$ is position as a function of time; $v(t)$ is velocity as a function of time; $a(t)$ is acceleration as a function of time. As a consistency check, you can verify that taking time derivatives of the $x(t)$ equation gives you the $v(t)$ equation, and taking time derivatives of the $v(t)$ equation gives you the $a(t)$ equation.

Focusing on the first equation, if we have values for all-but-one of the variables, we can "invert" the equation to solve for the remaining value. In your problem it looks like we have $$t=64, x(0)=2, x(t)=10$$ So $$ 10 = 2 + v(0)(64) + \frac{a}{2}(64)^2$$ We cannot solve for $a$ unless you specify $v(0)$. In my comment above I explicitly stated that was 0 (to specify a meaningful example problem).

Answer 2

You may be looking for something like $$t_1-t_0 :\min_{t_1}\left\{\int_{t_0}^{t_1} f'(t)dt=f_s-f_i\right\} \text{ if } f(t_0) = f_i$$

• $t_0$ initial time
• $t_1$ time for specific value
• $f_i$ initial value at $t_0$
• $f_s$ specific value to reach

It is an integral equation on the time-derivative of the variable/function you are considering.