Edit: This question involves derivatives, please read my prior work!
This question has me stumped.
A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant deceleration, find the value of deceleration that accomplishes this.
First, from the instructions, I believe…
$f'(x)=(5280/60)-ax=88-ax$
$f''(x)=-a$
I also believe $f(x)={\int}f'(x)dx=88x-a{\int}x=88x-a\frac{x^2}{2}$, because a is known to be constant.
Where I'm lost is what comes next. I can compute a and x in terms of each other at f(x)=450, but this doesn't seem to get me closer to the answer. Neither does the fact that $f^{-1}(450)=x$. What am I missing here? Thank you!
Best Answer
$a = \frac {dv}{dt}$ and $v = \frac {dx}{dt}$
By the chain rule $a = \frac {dv}{dx}\frac {dx}{dt} = v\frac {dv}{dx}$
$\int a\ dx = \int v\ dv\\ ax = \frac 12 v^2\\ a = \frac {v^2}{2x}$
But this is the minimum value of $a.$
$a \ge \frac {88^2}{900}$
Alternative....
$a = \frac {dv}{dt}\\ v = \frac {dx}{dt}$
$v(t) = v_0 - at = 88 - at\\ x(t) = 88 t - \frac 12 a t^2$
The vehicle comes to a stop at time $t = t^*$
$v(t^*) = 0\\ t^* = \frac {88}{a}\\ x(t^*) = 88\cdot \frac {88}{a} - \frac 12 a \frac {(88)^2}{a^2} = \frac {(88)^2}{2a} < 450\\ a > \frac {88^2}{900}$