Question

1. An airplane accelerates down a runway at 3.20 m/s2 for 3 s until is finally lifts off the ground. Determine the distance travelled before take-off​

Answer 1

Answer:

the correct answer is 0 because it's final velocity is 0 so

Answer 2

Answer:

s= 14.4 m

Explanation:

Given

$a = {3.20m} {s}^{ - 2} \\ t = 3 sec$

to find

$s(distance) =$

Now,

here initial velocity (u) =0

$a = \frac{(v - u)}{t} \\ 3.2 = \frac{(v - 0)}{3} \\ v = 3.2 \times 3 \\ v = 9.6m {s}^{ - 1}$

$s = ut + \frac{1}{2} a {t}^{2} \\ s = 0 \times 3 + \frac{1}{2} \times 3.2 \times {3}^{2} \\ s = 1.6 \times {3}^{2} \\ s = 14.4m$