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Time Loses Integrity

Problem

The velocity of a body increases with constant acceleration. Its motion is recorded between two points, $A$ and $B$, that are one metre apart, with $M$ being the midpoint of $AB$.

The body takes $p$ seconds to travel from $A$ to $M$ and $q$ seconds to travel from $M$ to $B$.

Given that the change in speed from $A$ to $B$, measured in metres per second, is integer, prove that $q$ cannot be integer.


Solution

Let $a$ represent the constant acceleration, $u$ represents the initial velocity as it passes $A$, and $v$ represents the final velocity as it passes $B$. Although we know that $AM = MB = 0.5$, we shall solve the general case initially, and let $AM = MB = d$.

Consider the following diagram.


We shall prove this in two different ways.

First proof:
Using the equation of motion, $s = ut + 0.5at^2$.

From A to M:

$$d = up + 0.5ap^2 \ \ \ \ [1]$$

From A to B:

$$\begin{align}2d &= u(p + q) + 0.5a(p + q)^2\\&= up + 0.5ap^2 + uq + 0.5a(q^2 + 2pq)\\&= d + uq + 0.5a(q^2 + 2pq)\\\therefore d &= uq + 0.5aq(q + p) \ \ \ \ \ \ \ \ \ [2]\end{align}$$

Multiplying equation $[1]$ through by $q$ and equation $[2]$ through by $p$ gives the following.

$$dq = upq + 0.5ap^2 q \implies upq = dq - 0.5ap^2 q$$$$dp = upq + 0.5apq(2p + q) \implies upq = dp - 0.5apq(2p + q)$$

$\begin{align}\therefore dp - 0.5apq(2p + q) &= dp - 0.5ap^2 q\\dp - dq &= 0.5apq(2p + q) - 0.5ap^2 q\\d(p - q) &= 0.5apq(2p + q - p)\\&= 0.5apq(p + q)\\2d(p - q) &= apq(p + q)\\\therefore a &= \dfrac{2d(p - q)}{pq(p + q)}\end{align}$

But as $v = u + at$, the change in speed, $c = v - u = at = a(p + q)$.

$\therefore c = \dfrac{2d(p - q)}{pq(p + q)} \times (p + q) = \dfrac{2d(p - q)}{pq} = \dfrac{p - q}{pq}$ (as $d = 0.5 metres)

$\begin{align}\therefore cpq &= p - q\\cpq + q &= p\\q(cp + 1) &=p\end{align}$

Hence the time taken to travel from M to B, $q = \dfrac{p}{cp + 1}$.

We were told that the body is increasing in speed, so $c$ is a positive integer. Therefore $cp + 1 \gt p$, and we prove that $q$ cannot be integer.

Alternative proof:
Let $T = p + q$. As $c = v - u = aT$, the constant acceleration, $a = \dfrac{c}{T}$.

Using $s = ut + 0.5at^2$, we get $1 = uT + 0.5\dfrac{c}{T}T^2 = T(u + 0.5c)$.

Hence $T = \dfrac{1}{u + 0.5c}$.

Clearly $T$ is maximised when the denominator is minimised; that is, $u = 0$ and $c = 1$, and we get $T = 2$ seconds.

As this is the maximum time the body can take to travel from $A$ to $B$ and speed is increasing, it must take less than 1 second to pass from $M$ to $B$, and cannot be integer.

Extension:
We note from the first proof that $cpq = p - q$, and this can rearranged to give the equation, $p = \dfrac{q}{1 - cq}$. If we let $c = 2$ and let $q = 0.4$, we get $p = 2$, which means that $T = 2.4$ seconds.

How can this be so?
Explain why $p$ cannot be integer.
What about the total length of time to travel from $A$ to $B$?

Problem ID: 272 (16 Feb 2006)     Difficulty: 4 Star

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