An airplane sitting at rest starts to accelerate in the east direction at [tex]2.38 \, \text{m/s}^2[/tex]. The plane travels [tex]2,400 \, \text{m}[/tex] before taking off.

How much time passes before the plane takes off?

[tex] t = [?] \, \text{s} [/tex]



Answer :

To determine the time [tex]\( t \)[/tex] it takes for an airplane to travel [tex]\( 2400 \, \text{m} \)[/tex] with a constant acceleration of [tex]\( 2.38 \, \text{m/s}^2 \)[/tex], we can use a kinematic equation of motion.

Given:

- The initial velocity [tex]\( u = 0 \, \text{m/s} \)[/tex], since the plane starts from rest.
- The acceleration [tex]\( a = 2.38 \, \text{m/s}^2 \)[/tex].
- The distance [tex]\( s = 2400 \, \text{m} \)[/tex].

We use the kinematic equation:

[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]

Since [tex]\( u = 0 \)[/tex]:

[tex]\[ s = \frac{1}{2}at^2 \][/tex]

Rearranging this equation to solve for [tex]\( t \)[/tex]:

[tex]\[ t^2 = \frac{2s}{a} \][/tex]

[tex]\[ t = \sqrt{\frac{2s}{a}} \][/tex]

Substitute the known values into the equation:

[tex]\[ t = \sqrt{\frac{2 \times 2400}{2.38}} \][/tex]

[tex]\[ t = \sqrt{\frac{4800}{2.38}} \][/tex]

[tex]\[ t \approx \sqrt{2016.8074} \][/tex]

[tex]\[ t \approx 44.908871313907184 \][/tex]

Therefore, the time [tex]\( t \)[/tex] it takes for the plane to travel [tex]\( 2400 \, \text{m} \)[/tex] before taking off is approximately [tex]\( 44.91 \, \text{s} \)[/tex].