Question

An airplane accelerates down a runmway at 2.9 m/s/s for 38.5 seconds

Answer 1

Given,

the initial velocity = 0 m /s.

acceleration = 3.20 m / s^2

time = 32.8 s

According to laws of motion.

s = ut + 1/2 at ^2

s = 1/2 at²

s=1/2(3.20)(32.8)²

s= 1721.344 m

the distance traveled before takeoff is 1731.3m

Answer 2

• initial velocity=u=0m/s^2
• Acceleration=2.9m/s^2
• Time=t=38.5s

Distance be s

$\\ \sf\longmapsto s=ut+\dfrac{1}{2}at^2$

$\\ \sf\longmapsto s=0(38.5)+\dfrac{1}{2}(2.9)(38.5)^2$

$\\ \sf\longmapsto s=1.45(1482.25)$

$\\ \sf\longmapsto s=2149.26m$