Question

An airplane accelerates down a runway at 3.20m/s^2 for 32.8s until is finally lifts off the ground.
Determine the distance traveled before take off. __________________________________WHO KNOW ONLINE THEY WILL GIVE ANSWER. IF DON'T KNOW PLEASE DON'T GIVE THE ANSWER.__________________ THANK YOU

Answer 1

Howdy!!

your answer is ----

Given,

acceleration a = -3.20m/s^2

time t = 32.8s

since, its final velocity v = 0

so,according to formula

→$v = u + at \\ = > 0 = u - 3.2 \times 32.8 = 104.96 \\ = > u = 104.96m {s}^{ - 1}$
so, its initial velocity u = 104.96m/s

Let, take off distance is s

so,
→ $s = ut + \frac{1}{2} a {t}^{2} \\ = > s = 104.96 \times 32.8 + \frac{1}{2} \times - 3.2 \times 32.8 \times 32.8 \\ = > s = 3442.668 - 1721.344 \\ = > s = 1721.324meter$

hence, take off distance is 1721.324meter

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hope it help you