solving a calculus problem without the area-under-a-graph approach I am working through Kline's calculus book and am absolutely stumped by a certain problem (3.23). I have googled up solutions to it using the area-under-the-curve approach, but the book has not discussed graphs in the context of derivative and integrals yet, so they must be looking for more of an algebraic solution. Here's the problem:
A subway train travel over a distance (s) over (t) seconds. It starts from rest and ends at rest. At the first part of its journey, it moves with a constant acceleration (f) and in the second, with a constant negative acceleration (r). Show that s = [fr / (f+r)] t^2 / 2
I've tried working out the formula starting with acceleration being (f), then speed being (ft), and position (f * t^2 / 2), and using (x) to denote the point in time when acceleration becomes negative, and ended up with a formula that's similar to the one being asked for, but not an exact match. I have now exhausted my ideas on how to approach the problem, and would appreciate some help in deriving the correct answer without the use of graphs.
EDIT: According to the expected solution, we start off by looking at the first part of the journey, with acceleration a = f, velocity v = ft and position s = ft^2 / 2 . So far, so good. Then they look at the second part, with a = -r (deceleration), and v = -rt + C. In determining C, it is suggested that if we treat the length of the first part of the journey as x, then when t = x, v = fx, and C = (f + r) * x. This is where I get lost - why is r (the negative acceleration part of the second part of the journey) a part of the constant? Shouldn't C only relate the the first part of the journey - the accumulated, starting speed from which we are now decelerating? And even if somehow the presence of r is justified here, why is it positive?
They then go on with stating that
v = -rt + (f+r)x , and
s = -rt^2 / 2 + (f+r)xt + C.
When t = x, s = fx^2 / 2 . Then
C = -[t^2 / 2] * (f+r) [I am not understanding where this is coming from either - why is f suddenly replaced by (f+r)?]
When trip ends, V = 0 or
-rt + (f+r)x = 0
Now that t is specified
x = rt / (t+r)  [I did obtain the same result in my attempts as well, but by writing v = -r(t-x) + fx = -rt + rx + fx => if v=0, tr = rx + fx => x = tr/(f+r); r is negative because we are decelerating and (t-x) denotes amount of time elapsed since we started decelerating, with t being total trip time and x being time when deceleration started]
Then substituting t for x, they get
s = [fr/(f+r)]t^2 / 2
Overall, their solution seems much simpler then either my attempts or the solutions presented here, but I am having difficulty following the logic of it all the way through. I've contacted the publisher for permission to post a picture of their solution here.
 A: The following figure

represents, on the same graphics 2 different curves (this is why the ordinate axis has no ticks):

*

*the distance as a function of elapsed time; its curve (in red) is the union of two parabolic arcs with a common speed at their junction (red circle).


*its derivative, the speed function ; its curve (in blue) is the union of two line segments meeting in a common point (blue circle) ; expressing the coordinates of this common point in two ways gives the following relationship :
$$T_1=\dfrac{r}{r-f}T_2\tag{1}$$
Besides, splitting the total travelled distance $s$ into

*

*the distance $s_1$ travelled during the acceleration phase, i.e.

$$fT_1^2/2\tag{3}$$

*

*the distance $s_2$ travelled during the deceleration phase, i.e.

$$-r(T_2-T_1)^2/2$$
(I have use for that formula (3) with elapsed time $T_1$ replaced by $(T_2-T_1)$ by symmetry); please note that $r$ being $<0$, we have taken $|r|=-r$),
gives
$$s=s_1+s_2=fT_1^2/2-r(T_2-T_1)^2/2\tag{2}$$
Taking (1) into account in (2) gives :
$$s=\dfrac{-fr}{f-r}\dfrac{T_2^2}{2}$$
in agreement with the formula you mention under the condition that $r$ is replaced by $-r$, i.e., taken with a positive sign.
Remark: the curves have been generated (with Matlab) by the following program
  f=2;r=-3;
  t2=10;t1=r*t2/(r-f);
  t=0:0.01:10;
  g=min(f*t,r*(t-t2));
  plot(t,g);
  plot(t,3*cumsum(g)/1000,'r')

A: Can you use some standard kinematics equations?
$v(t) = at\\
s(t) = \frac 12 at^2  + v_0 t$
$v_0 = 0$
While the train is accelerating
$s(t) = \frac 12 f t^2\\
v(t) = ft$
until some time $t = \tau$
For the back half of the journey....
We will be using $t$ for the time after $\tau$ for the rest of the work.
$v(t + \tau) = v(\tau) - rt\\
s(\tau + t) - s(\tau) = v(\tau)t - \frac 12 rt^2\\
s(\tau + t) = \frac 12 f\tau^2 + (f\tau)t - \frac 12 rt^2$
The trip ends when $v(t + \tau) = 0$
$v(t + \tau) = v(\tau) - rt = 0\\
f\tau - rt = 0\\
t = \frac fr\tau$
We will substitute for $t$ into $s(\tau + t) = \frac 12 f\tau^2 + (f\tau)t - \frac 12 rt^2$ from above.
$s(\tau + \frac fr\tau) = \frac 12 f\tau^2 + (f\tau)\frac fr\tau - \frac 12 r(\frac fr\tau)^2\\
s(\tau + \frac fr\tau) = \frac 12 f\tau^2 + \frac {f^2\tau^2}{r}  - \frac 12 (\frac {f^2\tau^2}{r})\\
s(\frac {f+r}r\tau) = \frac 12 f\tau^2(1+\frac {f}{r})$
The total time will be $t^* = \tau +  {f}{r}\tau = \frac {r+f}{r}\tau$
Substitute $\tau = \frac {r}{r+f}t^*$
$s(t^*) = \frac {fr}{2(f+r)}t^{*2}$
A: Note that the "constant negative acceleration" in the second part of the journey is not a negative number  $r$, but is $-r$ with $r>0$.
The journey of length $t$ is partitioned into two parts of time lengths $t_1$, $t_2$. We then have
$$t_1+t_2=t,\qquad ft_1=rt_2 \quad(=v_\max)\ .$$
Solving these equations for $t_1$ and $t_2$ gives
$$t_1={r\over f+r}\,t,\qquad t_2={f\over f+r}\,t\ .\tag{1}$$
On the other hand: With initial velocity $=0$ and constant acceleration $f>0$ (resp. $r>0$) you travel $f{t_1^2\over2}$ in time $t_1$, and  you travel $r{t_2^2\over2}$ in time $t_2$. Using symmetry with respect to $t\leftrightarrow -t$ and $r\leftrightarrow -r$ it follows that
$$f{t_1^2\over2}+r{t_2^2\over2}=s\ .$$
Inserting $(1)$ here we obtain
$$s=\left(f{r^2\over(f+r)^2}+r{f^2\over(f+r)^2}\right){t^2\over2}={fr\over f+r}\,{t^2\over2}\ .$$
A: The most concise solution use the suvat equations $s=\frac{v^2-u^2}{2a},\,t=\frac{v-u}{a}$ for the two constant-acceleration legs. Let $V$ denote the greatest speed, achieved before changing the acceleration from $f$ to $-r$. The distance travelled is $s=\frac{V^2-0^2}{2f}+\frac{0^2-V^2}{-2r}=\frac{V^2}{2}(1/f+1/r)$. Similarly, $t=V/f+(-V)/(-r)=V(1/f+1/r)$. Comparing these,$$s=\frac{Vt}{2}=\frac{t^2}{2(1/f+1/r)}=\frac{frt^2}{2(f+r)}.$$In particular, no area- or diagram-based arguments were needed to prove $S=\frac{Vt}{2}$.
A: I will just try to fill in some details of the book's solution.
But first, a review of the names of the parameters.

A subway train travel over a distance ($s$) over ($t$) seconds. It starts from rest and ends at rest. At the first part of its journey, it moves with a constant acceleration ($f$) and in the second, with a constant negative acceleration ($r$).

The solution then proceeds (paraphrased):

we start off by looking at the first part of the journey, with acceleration $a = f,$ velocity $v = ft$ and position $s = \frac12 ft^2.$

This is easy to understand, and also at odds with the problem statement, where $t$ was defined to be the total trip time. The velocity cannot be $ft$ at any point during the first part of the trip, because the elapsed time during that period is always less than $t,$ the total elapsed time at the end of the trip.
Let's follow the practice that the names in the problem statement are "official." So if there is a naming conflict in the solution, the conflicting name in the solution must change. I'll use the Greek letter $\tau$ as the name of a variable representing any elapsed time in the range from $\tau=0$ to $\tau=t.$
With that change, the solution now says that during the accelerating part of the trip,

velocity $v = f\tau$ and position $s =\frac12 f\tau^2 .$

Proceeding ahead, using $\tau$ variable elapsed time,

Then they look at the second part, with $a = -r$ (deceleration),

OK, again a little confusion: the problem statement says $r$ is negative acceleration, but apparently the numeric value of $r$ is positive.

and $v = -r\tau + C.$ In determining $C,$ it is suggested that if we treat the length of the first part of the journey as $x,$ then when $\tau = x,$ $v = fx,$

So at $\tau=x,$ (at the instant when we change from acceleration to deceleration),
$v = fx$ because we have been accelerating at a rate $f$ for $x$ seconds.
But we also just said that $v = -r\tau + C$ during the second part of the trip,
including the instant when the second part starts, namely when $\tau = x.$
Plugging $\tau = x$ into $v = -r\tau + C$, we get $v = -rx + C$.
So now we have two ways to calculate the velocity at time $\tau=x.$
Since the train can have only one velocity at that time, the two ways must produce the exact same answer, that is,
$$ -rx + C = fx $$
with the first formula on the right and the second on the left.
Now solve for $C$:
$$ C = fx + rx = (f+r)x. $$
It's just plain algebra. But an intuitive reason why $r$ shows up in the constant is that the term $-r\tau$ causes a problem that needs to be fixed: at time $\tau = x,$
we have just started decelerating and have not actually decelerated at all yet,
so the deceleration rate $-r$ should not yet have affected the velocity.
But since $\tau = x,$ the term $-r\tau$ comes out to $-rx$, which is not zero.
In order to stop this from unbalancing the formulas, we need to cancel this term by adding $rx$ back into the formula somewhere.

They then go on with stating that
$v = -r\tau + (f+r)x ,$

This is the formula for velocity during the deceleration part of the trip,
$v = -r\tau + C,$ with the value of the constant $C$ that was just calculated.

and $s = -\frac12 r\tau^2 + (f+r)x\tau + C.$

This reuses yet another variable name: $C$ was already used as the name of the constant in $v = -rt + C$ and now is being used as the name of a different constant.
Let's use a different name:

and $s = -\frac12 r\tau^2 + (f+r)x\tau + C_1.$

So that's the formula for the position during the deceleration part of the trip.

When $\tau = x,$ $s = \frac12 fx^2 .$

This is because $\tau = x$ occurs at the end of the acceleration, so the position must be $s =\frac12 f\tau^2 .$
But again the train can have only one position at this time, so the formula for the position during deceleration has to come up with the same answer at the start of the deceleration as we got at the end of the acceleration.
When  $\tau = x$ the formula for the deceleration says
$s = -\frac12 rx^2 + (f+r)x^2 + C_1,$ so
$$ -\frac12 rx^2 + (f+r)x^2 + C_1 = \frac12 fx^2 .$$
Solve for $C_1$:
$$ C_1 = \frac12 fx^2 - \left(-\frac12 rx^2 + (f+r)x^2\right)
= -\frac12  (f+r)x^2. $$
Note: the constant is not $-\left[\frac12 t^2 \right] (f+r).$
It may be calculated using the square of the elapsed time $x$ to the end of the acceleration, but not the elapsed time $t$ at the end of the trip nor any other time.
Again, an intuitive reason why we need both $f$ and $r$ in the constant here is because the formula for $s$ during the deceleration has terms with $r$ that need to be canceled out (there should not be any accumulated effect of the deceleration $r$ already at the very start of deceleration), and it also has "too much $f$" ($fx^2$ when it should be only $\frac12 fx^2$).
So now the complete formula for the distance at any time during the deceleration part of the trip is
$$ s = -\frac12 r\tau^2 + (f+r)x\tau - \frac12  (f+r)x^2. $$

When trip ends, $v = 0$ or $-rt + (f+r)x = 0.$

Solving the last equation for $x,$
$$ x = \frac{rt}{f+r}, $$
which you can plug into the equation for the position during deceleration,
\begin{align}
 s &= -\frac12 r\tau^2 + (f+r) \left( \frac{rt}{f+r} \right) \tau
     - \frac12 (f+r)\left( \frac{rt}{f+r} \right)^2 \\
 &= -\frac12 r\tau^2 +  (rt) \tau
     - \frac12 \left( \frac{r^2}{f+r} \right) t^2.
\end{align}
At the end of the trip, $\tau = t,$ so we plug that in:
\begin{align}
 s &= -\frac12 rt^2 +  rt^2 - \frac12 \left( \frac{r^2}{f+r} \right) t^2 \\
  &= \frac12 rt^2 - \frac12 \left( \frac{r^2}{f+r} \right) t^2 \\
  &= \frac12  \left( r - \frac{r^2}{f+r} \right) t^2 \\
  &= \frac12  \left( \frac{ (f+r) r - r^2}{f+r} \right) t^2 \\
  &= \frac12  \left( \frac{fr}{f+r} \right) t^2.
\end{align}
If the book's solution really did use $t$ for two different things, then it would have been impossible to spell out the steps in this much detail.
This seems very confusing (and not technically correct) to me.
