## The First Problem

A car has to travel a distance $s$ on a straight road. The car has maximum acceleration $a$ and maximum deceleration $b$. It starts and ends at rest.

Show that if there is no speed limit, the time is given by

$\displaystyle \sqrt{2s\cdot \frac{(a+b)}{ab}}$.

Solution: We are going to use two pieces of information:

• the area under the velocity-time graph is the distance travelled,
• the slope of the velocity-time graph is the acceleration.

First, a rough sketch of the velocity-time graph:

The slope of the first line segment is $a$, while the slope of the second line segment is $b$. The area under the curve is given by $s$, the total distance. The base is $t_1+t_2=T$, the total time, and the height is the maximum speed.

Let $v$ be the maximum speed reached. Using the area of a triangle, where $T=t_1+t_2$ is the total time:

$\displaystyle s=\frac12 Tv \Rightarrow v=\frac{2s}{T}$.

Now using the fact that the slope of the line segments, rise/run, is equal to the accelerations, we derive:

$\displaystyle a=\frac{v}{t_1}\Rightarrow t_1=\frac{v}{a}$, and

$\displaystyle b=\frac{v}{t_2}\Rightarrow t_2=\frac{v}{b}$.

Thus

$\displaystyle T=t_1+t_2=\frac{v}{a}+\frac{v}{b}=\frac{va}{ab}+\frac{vb}{ab}=v\cdot \frac{(a+b)}{ab}$.

Now, using $v=\frac{2s}{T}$:

$\displaystyle T=\frac{2s}{T}\frac{(a+b)}{ab}$

$\displaystyle \underset{\times T}{\Rightarrow } T^2=2s\cdot \frac{(a+b)}{ab}$.

Taking square roots (and noting $T>0$) completes the proof.

## The Second Problem (1997)

A particle, moving in a straight line, accelerates uniformly from rest to a speed $v$. It continues at this constant speed for a time and then decelerates uniformly to rest, the magnitude of the deceleration being twice that of the acceleration. The distance traveled while accelerating is 6 m. The total distance traveled is 30 m and the total time taken is 6 s.

i. Draw a velocity-time graph and hence, or otherwise, find $v$.

ii. Calculate the distance travelled at constant velocity.

Solution:

i. First the velocity-time graph:

There are number of things we need to use:

• the area under the velocity-time graph is the distance traveled
• the slope of the velocity-time graph is the acceleration

The question gives the following:

• the total distance is 30 m
• the total time is 6 s
• the distance traveled from $t=0$ to $t=t_1$ is 6 m.

In the first triangle:

$\displaystyle a=\frac{v}{t_1}\Rightarrow t_1=\frac{v}{a}$.

In the second triangle:

$\displaystyle 2a=\frac{v}{t_2}\Rightarrow t_2=\frac{v}{2a}=\frac{1}{2}t_1$,

and also $t_1=2t_2$. Define Let $t:=t_2$ so that $t_1=2t$. Therefore, using the fact that the total time is 6 s:

$t_1+t_2+t_3=2t+t_2+t=6\Rightarrow t_2=6-3t$.

Now, using the fact that the area under the first triangle is 6:

$\frac{1}{2}t_1v=tv=6$,

and the total area is 30:

$6+vt_2+\frac12 tv=30$

$\Rightarrow 6+v(6-3t)+3=30$

$\Rightarrow 6+6v-3\underbrace{tv}_{=6}+3=30$

$\Rightarrow 6v=30-6+18-3=39\Rightarrow v=\frac{13}{2}$ m/s.

ii. Using $tv=6$ we have $t=\frac{12}{13}$ so that the time spent at constant velocity is:

$\displaystyle t_2=6-3t=6-\frac{36}{13}=\frac{42}{13}$,

and therefore the distance traveled is:

$s_2=\frac{13}{2}\times \frac{42}{13}=21 \text{ m}$.

Alternatively note that $s_3=\frac12 s_1=3$ and so $s_2=30-s_1-s_2=21$ m.