How long does an object moving with positive uniform acceleration travelling distance $s$ for time $t$ move for the last $\frac{3}{4}$ portion?

Question

An object which moves with a positive value of uniform acceleration travels distance $s$ for time $t$. What is the time does the object move for the last $\dfrac{3}{4}$ portion of the aforementioned distance?

Let $s_0 = 0$, since knowing the initial displacement from the origin is irrelevant.

We have that $s = \dfrac{(v_0 + v)t}{2}$, $s' = \dfrac{(v_0 + v')t'}{2}$ and $s = 4s'$

(where $v$ and $v'$ are the velocities at time $t$ and $t'$, when the object reaches distances $s$ and $s'$).

$$\iff (v_0 + v)t = 4(v_0 + v')t' \iff v_0(4t' - t) + (4v't' - vt) = 0 \iff v_0 = -\dfrac{4v't' - vt}{4t' - t}$$

Plugging the previous equation, we have that $s = 4s' = \dfrac{2(v - v')tt'}{4t' - t}$.

Furthermore, it is true that $v = v_0 + at$ and $v' = v_0 + at' \implies v - v' = a(t - t')$

(where $a$ is the object's acceleration).

$\implies s = 4s' = \dfrac{2a(t - t')tt'}{4t' - t}$

And that's all I could do.

Answer 1

There are a couple of incorrect formulas in your proof. In the line starting "Furthermore, it is true..." the two formulas should either be $v=v_0+a\underline{t}$ and $v'=v_0+a\underline{t'}$ , or $v^2=v_0^2+2as$ and $(v')^2=v_0^2+2as'$, I'm not certain which ones you meant to use. Either way, though, you won't be able to continue as you did.

Your question seems to be asking for a solution in terms of $s$ and $t$ only, but this is impossible as the answer will depend on $v_0$ as well. As a simple example, consider the following two cases: (Assume usual metric units, i.e. m, s, m/s, etc.)

Case A:    $v_0=0, t=10, s=160$.  (This yields $v=32$ and $a=3.2$.)
                  In this case, the first ${\frac{1}{4}}^{\text{th}}$ of the distance is covered in $5$ seconds, so the final ${\frac{3}{4}}^{\text{ths}}$ of the
                  distance will also take $5$ seconds, i.e. $\frac{1}{2}t$.

Case B:    $v_0=6, t=10, s=160$.  (This yields $v=26$ and $a=2$.)
                  In this case, the first ${\frac{1}{4}}^{\text{th}}$ of the distance is covered in $4$ seconds, so the final ${\frac{3}{4}}^{\text{ths}}$ of the
                  distance will take $6$ seconds, i.e. $\frac{3}{5}t$.

Thus, even though $s$ and $t$ are equal in the above example, the fraction of $t$ required for the final ${\frac{3}{4}}^{\text{ths}}$ of the distance is different because of the differing initial velocities.

If we assume that $v_0=0$, then there's a simple solution. We use the equation $s=v_0 t + \frac{1}{2}at^2$ and get the system

$\quad s=\frac{1}{2}at^2\\ \quad s'=\frac{1}{2}a(t')^2$

Then, as $s=4s'$, we get $t' = \frac{1}{2}t$, so the time taken to cover the last ${\frac{3}{4}}^{\text{ths}}$ of the distance (lets call this $t^*$) will also be $\frac{1}{2}t$.

Unfortunately it gets messier if $v_0 \neq 0$. Again taking the equation $s=v_0 t + \frac{1}{2}at^2$, now we get

$\quad s'= \frac{1}{4}s = v_0 t' + \frac{1}{2}a(t')^2$

This yields the equation $2a(t')^2 + 4v_0t' - s = 0$. Applying the quadratic formula, a little simplifying and keeping only the positive root yields

$\quad t'= \frac{-2v_0 + \sqrt{4v_0^2 + 2as}}{2a}$

which then gives

$\begin{align} \quad t^* &= t-t'\\ &= \frac{2(at+v_0)-\sqrt{4v_0^2+2as}}{2a}\\ \end{align}$.

This does require $a$ which we can find by using the two formulas $s=\frac{v_0+v}{2}t$ and $v=v_0+at$; solving the first for $v$, plugging into the second and then solving for a yields

$\quad a=\frac{2(s-v_0t)}{t^2}$

At this point one COULD plug this formula for $a$ into the above quadratic formula, but this does not yield a particularly nice result.

So, to recap: if $v_0=0$, then $t^*=\frac{1}{2}t$; if $v_0 \neq 0$, then we use the pair of formulas

$\quad a=\frac{2(s-v_0t)}{t^2}$

followed by

$\quad t^* = \frac{2(at+v_0)-\sqrt{4v_0^2+2as}}{2a}$.

Answer 2

$s = ut + \dfrac{at^2}{2}$. Since nothing is mentioned about the initial velocity $u$ we assume that the object started from rest and $u = 0$. Hence $a = \dfrac{2s}{t^2}$.

Let $s_1$ and $s_2$ be the distance covered in time $t_1$ and $t_2$ starting from rest. Then,

$s_2 - s_1 = \dfrac{at_2^2}{2} - \dfrac{at_1^2}{2} = \dfrac{a}{2}(t_2^2 - t_1^2)$

Substituting $s_2 = s,t_2 = t, a = \dfrac{2s}{t^2}$ and $t_1 = t-\dfrac{3t}{4} = \dfrac{t}{4}$, we get

$s - s_1 = \dfrac{s}{t^2}(t^2 - t^2/16) = \dfrac{15s}{16}$.

Hence the object covers $\dfrac{15}{16}$-th of the total distance in the last $\dfrac{3}{4}$-th of the total travel duration.