How far 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 distance does the object move for the last $\dfrac{3}{4}$ portion of the aforementioned time?

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

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

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

$$ \iff \frac{s}{v_0 + v} = \frac{4s'}{v_0 + v'} \iff v_0(4s' - s) + (4s'v - sv') = 0 \iff v_0 = -\frac{4s'v - sv'}{4s' - s}$$

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

Furthermore, it is true that $v^2 = v_0^2 + 2as$ and $v'^2 = v_0^2 + 2as' \implies v' - v = \dfrac{2(s' - s)}{v + v'}$

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

$\implies t = 4t' = \dfrac{(4s' - s)(v' + v)}{s' - s}$

And that's all I could do.

Answer 1

The distance moved by an object with initial velocity $u_0$ and uniform acceleration $a$ in time $t$ is $$s=u_0t+\frac{1}{2}at^2$$

In this case you are given $s$ and $t$, but not $a$ or $u_0$ and you are asked to find the distance travelled between time $t/4$ and $t$. If you knew one of $a,u_0$ you could find the other. It is easier to take $a$ as the unknown rather than $u_0$, so we have $$u_0=\frac{2s-at^2}{2t}$$

Using the same formula for but with time $t/4$ you get for the distance $s'$ travelled between time 0 and time $t/4$:

$$s'=\frac{u_0t}{4}+\frac{1}{2}\ a\ \frac{t^2}{16}=\frac{8s-3at^2}{32}$$ So the distance travelled between $t/4$ and $t$ is $$\frac{24s+3at^2}{32}$$

Answer 2

We have that

$$s=v_0t+\frac12 at^2 \implies at^2+2v_0t-s=0 \implies t=\frac{-2v_0+\sqrt{4v_0^2+4as}}{2a}\implies t'=\frac34t$$

With the given data we can't obtain more than that.

Answer 3

Assume the object starts from rest. In the first $\frac {1}{4}$ of the time the object travels $\frac {1}{16}$th of the distance. (A nice way to see this is to consider the velocity-time graph.)

The answer you require is therefore $\frac {15}{16}s.$

Answer 4

We know that average velocity is $\frac s t$. If initial velocity is $u$ then final velocity is $\frac {2s}{t} - u$. If acceleration is $a$ then we have

$u + at = \frac {2s}{t} - u\\\Rightarrow a=\frac {2s}{t^2} - \frac{2u}{t}$.

At time $\frac t 4$, velocity is $u + \frac{at}{4}$ so average velocity over the first $\frac 1 4$ of the time period is $u + \frac{at}{8} = \frac{s}{4t}+\frac{3u}{4}$ and the distance travelled during the first $\frac 1 4$ of the time period is

$(\frac{s}{4t}+\frac{3u}{4})\frac{t}{4}=\frac{s+3ut}{16}$

So the distance travelled during the last $\frac 3 4$ of the time period is

$ s- \left(\frac{s+3ut}{16}\right) = \frac{15s-3ut}{16}$

If we assume $u=0$ then this reduces to S. Dolan's answer above.