Question

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

Answer 1

Equation of motion :- s = ut + 1/2at²

Given,
Intial velocity = 0 m/s .
Acceleration = 3.20m/s²
Time = 32.8

Distance = 0*32.8 + (3.2)(32.8)²
= 1721.344 m

Therefore, The aeroplane travelled 1721.344 m before taking off.

Answer 2

Given- Initial velocity, u = 0 m/s
Time taken, t = 32.8 seconds
Acceleration, a = 3.20 m/s^2

To find - Distance travelled =?

Sol : By Second Equation of Motion

$S = ut + \frac{1}{2} at {}^{2} \\ \: \: \: \: = 0 \times 32.8 + \frac{1}{2} \times 3.20 \times (32.8) {}^{2} \\ \: \: \: \: = \frac{1}{2 } \times 3.20 \times (32.8) {}^{2} \\ \: \: \: \: = 1721.344 \: m$

Ans: Hence, the distance traveled by the airplane before taking off is 1721.344 metre.