Question

To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeoff velocity, depends on several factors, including the weight of the aircraft and the wind velocity.

part A plane accelerates from rest at a constant rate of 5.00m/s2 along a runway that is 1800m long. Assume that the plane reaches the required takeoff velocity at the end of the runway. What is the time tTO needed to take off?
part C What is the distance dfirst traveled by the plane in the first second of its run?
Part D What is the distance dlast traveled by the plane in the last second before taking off?

Answer 1

A) 26.84 s

First of all, we can find the velocity at which the plane takes off, using the equation

$v^2-u^2=2ad$

where

v is the take-off velocity

u = 0 is the initial velocity

$a=5.00 m/s^2$ is the acceleration

d = 1800 m is the distance covered

Solving for v,

$v=\sqrt{2ad}=\sqrt{2(5)(1800)}=134.2 m/s$

The time needed to reach this velocity can be found by using

$v=u+at$

And solving for t,

$t=\frac{v-u}{a}=\frac{134.2-0}{5}=26.84 s$

C) 2.5 m

The distance travelled by the plane in the first second of its run can be found by using

$d=u\Delta t+\frac{1}{2}a(\Delta t)^2$

where

u = 0

$\Delta t=1 s$ is the time interval

$a=5.00 m/s^2$ is the acceleration

Substituting,

$d=0+\frac{1}{2}5(1)^2=2.5 m$

D) 131.7 m

To solve this part, we need first to find the velocity of the plane one second before taking off, which is at

$t=26.84 -1 = 25.84 s$

The velocity can be found as follows:

$v=u+at=0+(5)(25.84)=129.2 m/s$

Now we can find the distance travelled by the plane during the last second by using

$d=v\Delta t+\frac{1}{2}a(\Delta t)^2$

where

v = 129.2 m/s

$\Delta t=1$

$a=5.00 m/s^2$ is the acceleration

Substituting,

$d=(129.2)(1)+\frac{1}{2}5(1)^2=131.7 m$