No Times Are Given!

No big formulas.

Draw the graph of velocity :

Average velocity is $\dfrac{S+2S+5S}{T+T+5T}=\dfrac{8S}{7T}$ . Max velocity is $\dfrac{2S}{T}$ . Ratio is $\boxed{\dfrac47}$ .

I have faith that there is a more elegant solution than mine, but I certainly had fun crunching the numbers and finally getting a right answer.

We start by making important assumptions that do not lose generality: $s(0) = 0$ , $t = t_1$ is the time at which the acceleration $a$ ends, $t = t_2$ is the time at which uniform deceleration $d$ begins, and $t = t_3$ is the time at which the body stops.

Let the body's initial acceleration be constant a. Then $v(t) = \int adt = at + C$ , but $C$ = 0 because the body starts from rest. Integrating again gives us position $s(t) = \int atdt = 0.5at^2 + C$ , but we can again say the value of $C$ is 0 because the body starts at the origin. Therefore, $s(t_1) = 0.5at_{1}^{2} = S$ . Moreover, the velocity of the body at this time will be $v(t_1) = at_1$ and it is the body's maximum velocity because it doesn't experience any more positive acceleration.

Now, over the interval $t_1 < t < t_2$ , the velocity is constant $v(t) = at_1$ and the position function will then be $s(t) = at_1t + C$ . Since it is known that $s(t_1) = 0.5at_{1}^{2}$ , the value of $C$ must be $-0.5at_{1}^{2}$ , giving the position function $s(t) = at_1t - 0.5at_{1}^{2}$ . At $t_2$ , the body has traveled a total of $3S$ , so $s(t_2) = at_1t_2 - 0.5at_{1}^{2} = 3(0.5at_{1}^{2}) \longrightarrow t_2 = 2t_1$

Lastly, over the interval $t_2 < t < t_3$ , the body decelerates at constant rate $d$ . Then $v(t) = \int ddt = dt + C$ , but since the body has velocity of $at_1$ from the previous interval, $C = at_1 - dt_2 = at_1 - 2dt_1$ so the velocity function is $v(t) = dt + at_1 - 2dt_1$ . Since $v(t_3) = 0$ , we have that $v(t_3) = dt_3 + at_1 - 2dt_1 = 0 \longrightarrow t_3 = \frac{2dt_1 - at_1}{d}$

To find a relationship between $a$ and $d$ , we use the physics formula $v_f^2 = v_i^2 + 2a\Delta x$ to find $0 = (at_1)^2 + 2(d)(5S) \longrightarrow -a^2t_1^2 = 5dat_1^2 \longrightarrow d = -0.2a$

Now to find:

$\dfrac{\text{average velocity}}{\text{max velocity}} = \dfrac{\dfrac{\text{displacement}}{\text{total time}}}{\text{max velocity}} = \dfrac{\frac{8S}{t_3}}{at_1} = \dfrac{8(0.5at_{1}^{2})}{t_3}\cdot \dfrac{1}{at_1} = \frac{4t_1}{ \frac{2dt_1 - at_1}{d}} =\frac{4d}{2d - a}$

Since $d = -0.2a$ , $\dfrac{-0.8a}{-0.4a - a} = \dfrac{4}{7}$ . $\square$

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As a bonus, I determined a position function for a body that adheres to the criteria set forth by the problem:

$s(t) = \begin{cases} t^2 & 0 < t < 1 \\ 2t - 1 & 1 < t < 2 \\ 0.2(-t^2 + 14t - 9)& 2 < t < 7 \end{cases}$

I just don't want to write down the entire solution, but a v-t graph definitely makes it very quick.

Lets put 3 time frames: - Acceleration $a_1$ , traveling a distance $S$ and time $t_1$ - Constant velocity, traveling distance $2S$ and time $t_2$ - Deceleration $a_2$ , traveling distance $5S$ and time $t_3$

We know that $s=\frac{at^2}{2} , s=vt$ and $a=\frac{\Delta v}{\Delta t}$ , then we could assume $v_{max}$ is the maximum velocity.

$\begin{cases} a_1=\frac{v_{max}}{t_1} , S=\frac{a_1t_1^2}{2} \rightarrow t_1=\frac{2S}{v_{max}}\\ 2S=v_{max}t_2 \rightarrow t_2=\frac{2S}{v_{max}}\\ a_2=\frac{v_{max}}{t_2} , 5S=\frac{a_2t_2^2}{2} \rightarrow t_3=\frac{10S}{v_{max}}\end{cases}$

$v_{avg}=\frac{S+2S+5S}{t_1+t_2+t_3}=\frac{8S}{\frac{2S+2S+10S}{v_{max}}}=\frac{8Sv_{max}}{14S}=\frac{8}{14}v_{max}$

So the ratio of $v_{avg}$ to $v_{max}$ is $\frac{v_{avg}}{v_{max}}=\boxed{\frac{4}{7}}$