Using kinematics we can find that the take-off distance is 6163 ft
Given parameters
- The initial and final speed of the plane i = 0 and v = 140 mph
To find
The measurement system allows not to have problems when working in different units, in this case we reduce the speed units
v = 140 mile / h (5280 ft / mile) (1h / 3600 s) = 205.34 ft / s
The kinematics allows to find the relationships between the position, the speed and the acceleration of a body, in this case the movement is in one dimension.
v = v₀ + a t
where v and v₀ are the final and initial velocity, respectively, at acceleration and t the time
a = [tex]\frac{v-v_o}{t}[/tex]
a = [tex]\frac{205.34 - 0}{60.0}[/tex]
a = 3.42 ft / s²
Let's use the expression
v² = v₀² + 2 a x
Where v and v₀ are the final and initial velocity, respectively, at acceleration and x the distance traveled
x = [tex]\frac{v^2 - v_o^2 }{2 \ a}[/tex]
x = [tex]\frac{205.33^2 - 0}{2 \ 3.42}[/tex]
x = 6163.8 ft
Let's reduce to miles
x = 6163.8 ft (1 mile / 5280 ft)
x = 1.17 mile
In conclusion using kinematics we can find that the take-off distance is 6163 ft
Learn more about kinematics here:
