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,
$0=-c\frac{1}{400}+100$
$c=40000$
$a=40000 miles/hour^2$
Best Answer
At, $s = \frac{1}{4}, v = 0$, find time taken to get to $v = 0$
So, $v = 0 = -ct+100, t = \frac{100}{c}$
$\frac{1}{4} = - \frac{c}{2} (\frac{100}{c})^2 + \frac {100^2}{c}$
$\frac{1}{4} = \frac {100^2}{2c}$
$c = 2 \times 100^2$