Question

The distance traveled by an airplane before it takes off if it starts from rest and accelerates down a runway at 3.50 m/s^2 for 34.5 s

Answer 1

Given,

the initial velocity = 0m/s.

acceralation = 3.20m /s^2

time = 32.8 s

According to the law of motion

s = ut + 1/2 at ^2

s = 1/2 at ^2

s = 1/2 (3.20)(32.8)^2

s = 1721.344 m

Answer 2

2082.9375m

Step-by-step explanation:

wkt s=ut+1/2at^2

u=o initial velocity is at rest

substitute value of a and t in the formula

s=0+1/2×3.5×34.5^2

s= 2082.9375m