Question

an airplane lands and starts down the runway at a northwest velocity of 55.7 m/s. what constant acceleration allows it to come to a stop in 1.24 km? (in m/s^2)

Answer 1

Solution:

$\upsilon=\upsilon_0-at$

But

$\upsilon=0$

So:

$\upsilon_0=at$

And than:

$t=\frac{\upsilon_0}{a}$

Distance before the stopping:

$S=\upsilon_0t-\frac{at^2}{2}=\frac{\upsilon^2_0}{a}-\frac{a}{2}\frac{\upsilon^2_0}{a^2}=\frac{\upsilon^2_0}{2a}$

Than we can find the acceleration:

$a=\frac{\upsilon^2_0}{2S}=\frac{(55.7)^2}{2\cdot{}1.24\cdot{}10^3}=1.25$ m/s².

Answer:

a = 1.25 m/s².