Question

Aircraft carriers are very large ships that launch jets. The length of an aircraft carrier is 150 m. A jet taking off accelerates uniformly from rest and travels the length of the aircraft carrier in 2.00 s. What is the magnitude of the jet’s velocity when it leaves the ship?

Answer 1

-- For the whole time that the jet is still on the ship,
its average speed is 75 meters per second.

-- Its average speed is (1/2) (start speed + end speed)

-- But its end speed is (2 x acceleration)
so
-- 75 = (1/2) (0 + 2·acceleration)

75 = (1/2) (2·acceleration)

acceleration = 75 m/s²

= about 7.6 G's ! !

Answer 2

`v_(f)=150 m/s`

Step-by-step explanation:

Using one of the uniformly acceleration movement.

`x=x_(i)+v_(i)t+(1/2)at^(2)`

We know that:

• x(i) initial position. In our case 0
• v(i) initial velocity. In our case 0

So, we can find a:

`x=(1/2)at^(2)`

`a=(2x)/(t^(2))`

`a=(2*150)/(2^(2))`

`a=75 m/s^(2)`

Now, let's use the acceleration definition:

`a=(\Delta v)/(\Delta t)`

`a=(v_(f)-v_(i))/(t)`

Therefore the velocity of the jet at the end of the aircraft will be:

`a=(v_(f))/(t)`

`v_(f)=at=75*2=150 m/s`

I hope it helps you!