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.
Best Answer
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 $\tau = \frac {r}{r+f}t^*$
$s(t^*) = \frac {fr}{2(f+r)}t^{*2}$