Question

An aeroplane has a run of 500 m to take
off from the runway. It starts from rest
and moves with constant acceleration to
cover the runway in 30 sec. What is the
velocity of the aeroplane at the take off ?

Answer 1

Answer:

33m/s.

Explanation:

Since, the time in which the aeroplane covers the runway is 30 seconds and the distance of the runway is 500 meter.

So, we know from newtons second law of motion that distance s = ut + 1/2at^2. Where we know that the initial velocity of the aeroplane was 0 m/s and the time taken is 30 sec.

So, the acceleration will be 1.1 m/s^2. So again applying the acceleration in Newton's first law of motion we will get v= u +at or the value of v will be 1.1*30 = 33 m/s. So the velocity of the plane at takeoff will be 33 m/s.

Answer 2

Answer:

Explanation:

Hope it helps...