The landing velocity of an airplane is 100 mi/hr. Constant deceleration and comes to a stop after traveling 1/4 mile. Find the deceleration.

Question

Please could someone confirm that my calculations and answer is correct.

The landing velocity of an airplane (i.e., the velocity at which it touches the ground) is 100 mi/hr. It decelerates at a constant rate and comes to a stop after traveling mile along a straight landing strip. Find the deceleration or the negative acceleration.

$a=-c$

$v=-ct+100$

$s=-\frac{1}{2}ct^2+100t$

at $\frac{1}{4} = s, -c=0$

$\frac{1}{400}=t$ at $s=\frac{1}{4}$

at $t=\frac{1}{400}$ $v=0$

Therefore,

$0=-c\frac{1}{400}+100$

$c=40000$

$a=40000 miles/hour^2$

Answer 1

At, $s = \frac{1}{4}, v = 0$, find time taken to get to $v = 0$

So, $v = 0 = -ct+100, t = \frac{100}{c}$

$\frac{1}{4} = - \frac{c}{2} (\frac{100}{c})^2 + \frac {100^2}{c}$

$\frac{1}{4} = \frac {100^2}{2c}$

$c = 2 \times 100^2$

Answer 2

I do not understand your line $\frac14=s,-c=0$. I understand why $s=1/4$, but why is $-c=0$ is unclear to me.

Here is how I would approach it. Assuming constant deceleration $a=-c$ and the terminal time is $T$, the first equation implies $$ v(t) = v_0 + at = 100 - ct \implies 0 = 100 - cT \iff T = \frac{100}{c}. $$ The second equation says $$ s(t) = 100t - \frac{ct^2}{2} $$ but we want this at terminal time $T$, then $$ \frac14 = s(T) = s(100/c) = 100 \frac{100}{c} - \frac{c}{2} \left(\frac{100}{c}\right)^2 = \frac{100^2}{2c}. $$ You can easily solve this for $c$, can you finish this?