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].
