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

To determine the time needed for an airplane to take off given the acceleration and runway length, kinematic equations are used. The time for takeoff is calculated as 24 seconds, and the distances traveled in the first and last seconds are determined using similar principles.

Step-by-step explanation:

To determine the takeoff time tTO for an airplane that accelerates from rest at a constant rate of 5.00 m/s2 along a runway that is 1800 m long, we can use the kinematic equation:

s = ut + ½ at2

where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. Given that the initial velocity u is 0 and the distance s is 1800 m:

1800 m = 0 + ½ × (5.00 m/s2) × t2

Solving for t, we get that the time tTO is 24 seconds.

For part C, the distance dfirst in the first second can be found with:

dfirst = ut + ½ at2

Substituting the given values, dfirst = 0 + ½ × (5.00 m/s2) × (1 s)2 = 2.5 m.

For part D, to find the distance dlast in the last second before takeoff, we first calculate the final velocity using v = u + at. Then we use the velocity one second before takeoff (v - a) to find the distance covered in that second using dlast = (v - a) + ½ × a × (1 s)2.

Answer 2

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=√(2ad)=√(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=(v-u)/(a)=(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+(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+(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+(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)+(1)/(2)5(1)^2=131.7 m`