Respuesta :
Given:
Final Velocity = 360 km/h
Acceleration, a = 3.60 x 10⁴ km/h²
Let's solve for the following:
• (a). How many kilometers long must the runway be?
To find the length of the runway, apply the formula:
[tex]v^2_f=v^2_i+2ad[/tex]Where:
vf is the final velocity = 360 km/h
vi is the initail velocity = 0 km/h
a is the acceleration = 3.60 x 10⁴ km/h²
d is the distance the plane will cover which in this case is the length of the runway.
Rewrite the equation for d:
[tex]d=\frac{v^2_f-v^2_i}{2a}[/tex]Input values in the equation:
[tex]\begin{gathered} d=\frac{360^2-0^2}{2(3.60\times10^4)} \\ \\ d=\frac{360^2}{2(3.60\times10^4)} \\ \\ d=\frac{129600}{72000} \\ \\ d=1.8\text{ km} \end{gathered}[/tex]Therefore, the runway must be 1.8 kilometers long.
• (b). How many seconds will a plane need to accelerate to take-off speed?
To find the number of seconds it will take, apply the formula:
[tex]v_f=v_i+at[/tex]Where t is the time.
Rewrite the equation for t:
[tex]t=\frac{v_f-v_i}{a}_{}[/tex]Thus, we have:
[tex]\begin{gathered} t=\frac{360-0}{3.60\times10^4} \\ \\ t=\frac{360}{3.60\times10^4} \\ \\ t=0.01 \end{gathered}[/tex]The time is 0.01 hour.
To convert to seconds, we have:
Where:
1 hour = 3600 seconds
0.01 hour = 3600 x 0.01 = 36 seconds
Therefore, the plane will need to accelerate for 36 seconds before take off.
ANSWER:
(a) 1.8 km
(b) 36 seconds
