Need Help...Physics Question?

Question

An airplane accelerates down a runway at 3.20m/s2 for 32.8s until its finally lifts off the ground. Determine the distance traveled before take off?


Im so lost at school atm can someone solve it by showing the work so i can review it and understand it....Greatly Appreciated thanks

Answer 1

Assuming constant acceleration from zero, final velocity is acceleration * time, and average velocity is 1/2 final velocity. Distance traveled is average velocity * time.

So:
3.20 m/s^2 * 32.8 s = 104.96 m/s
1/2 * 104.96 m/s = 52.48 m/s
52.48 m/s * 32.8 s = 1721.344 m

Since the data given indicates the answer should have 3 significant figures, the answers above would be 105 m/s , 52.5 m/s , and 1720 m.

Answer 2

s = 1/2 at^2
= 0.5 * (3.2) * (32.8)^2 ....... [units = (m/sec^2) * (sec^2) = m ]
= 0.5 * (3.2) * 1075.84 m
= 1.6 * 1075.84 m
= 1721.344 m

.........
where
s = "distance"
a = acceleration
t = time