Question

A plane accelerates from rest at a constant rate of 5.00 ms-2 along a runway that is 1800 m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time t needed to take off?

What is the speed v of the plane as it takes off?

Answer 1

The time needed to take off is 27 sec.

Speed of the plane as it takes off is 134.16 m/s

Since the plane accelerates with constant acceleration, therefore we can use the equation of motion .

The time needed to take of the plane can be calculated by the 2nd equation of motion

S=ut+0.5a$t^{2}$

Here, u is the initial velocity, t is time, S is distance

Plugging the values in the above equation

1800=0×t+0.5×5×$t^{2}$

t=26.83=27 sec

Now the velocity of the plane as it takes off can be calculated by the 1st equation of motion

v=u+at

here, v is the final velocity

Plugging the values in the above equation

v=0+5×26.83=134.16 m/s