Question

An airplane accelerates down a runway at 3.20 m/s for 32.8 s until is dinally lifts off the ground. How fast is the airplane after 32.8 s?

Answer 1

Answer:

Initial speed of airplane u=0 m/s

Acceleration of the plane a=3.2 m/s

2

and time taken t=3.28 s

Using formula, distance covered S=ut+

2

1

at

2

⇒ S=0(32.8)+0.5(3.2)(32.8)

2

=1721.3∼1720m