Airplane A , starting from rest with constant acceleration, requires a runway 400 m long to become airborne. Airplane B requires a takeoff speed twice as great as that of airplane A , but has the same acceleration, and both planes start from rest.

Questions

1. How long must the runway be for airplane B

2.If airplane A takes time T to travel the length of its runway, how long (in terms of T) will airplane B take to travel the length of its runway?

Explanation:

Constant acceleration question

Airplane A start from rest

Then, the initial velocity is u=0m/s.

The runaway before is flight is 400m.

Time A use to travel is T

Therefore, t1=T

For plane B

Airplane B starts from rest too

i.e u=0m/s.

Let the distance Plane B covers be L.

Let the time plane B use to travel be t

Given that they both have the same acceleration

Then,

The final speed of airplane B is twice that of plane A.

Let the final speed of plane A =x

Then, final speed of plane B = 2x.

Using equation of motion

V=u+at

V^2=U^2+2as

S=ut+1/2at^2

For plane A

V=u+at; x=0+aT,

since T is time plane A use to travel

x=aT

For plane B

V=u+at; 2x=0+at

Since t is the time plane B use

2x=at Equation 2

Then, substituting x into equation 2

2(aT)=at

Then

2aT=at.

Divide both side by a, since the both have the same acceleration

2T=t

Also using

V^2=U^2+2as

Plane A.

(x)^2=0^2+2(a)(400)

x^2=800a

Plane B

(2x)^2=0^2+2(a)(L)

4x^2=2aL

divide both side by 2

2x^2=aL

Now substituting x^2=800a

2(800a)=aL

1600a=aL

Since they both have the same acceleration, divide both side by a

Then,

L=1600m

1. The runway distance the plane B will cover is 1600m

2. And the time B will cover the runway in terms of time(T) A covers is run way.

t=2T

i.e twice the time plane A will cover it owns distance

the answer is c i did the same question so i know that's right!

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft