Question

Assuming a constant acceleration of $a_x=4.3m/s^2$ starting from rest, what is the airplane's takeoff velocity after 18.4 s? How far down the runway has the plane moved by the time it takes off?

• A. 760 m
• B. 728 m
• C. 740 m
• D. 750 m

Answer 1

Answer: (B)

728 m

After $t=18.4sec$ the velocity will be $v=a_{x}t=4.3\times 18.4=79.12m/s$

After $t=18.4sec$ the distance covered in this time will be $S=\frac{a_x{t}^2}{2}=\frac{4.3\times {18.4}^2}{2}=727.9m$