Question

A plane takes off from a runway and begins its climb. The plane takes off at a constant acceleration of 20 LaTeX: m /s^{2} . Once it reaches altitude it moves at 246 m/s, and maintains that speed. If it continues at this speed for 93 min, how far would the plane have traveled from the time took off?

Answer 1

Answer:

The plane have traveled 1512.9 m from the time took off

Step-by-step explanation:

Acceleration $a = 20m/s^2$

Initial speed = u = 0

Final speed = v = 246 m/s

Time = t = 93

We are supposed to find how far would the plane have traveled from the time took off

We will use equation of motion

$v^2=u^2+2as\\(246)^2=2\times 20 \times s\\\frac{(246)^2}{2 \times 20}=s\\1512.9 m=s$

Hence The plane have traveled 1512.9 m from the time took off