Question

PLEASE HELP: An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 140mph at take-off. If the plane can reach this take-off speed in 60.0s, how far will the plane travel before take-off. Assume the plane increases speed steadily. How far does the plane need to travel during take-off (in given units)?

Answer 1

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

• Time t = 60.0 s

To find

• The distance to take off

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 = $\frac{v-v_o}{t}$

          a = $\frac{205.34 - 0}{60.0}$

          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 = $\frac{v^2 - v_o^2 }{2 \ a}$

              x = $\frac{205.33^2 - 0}{2 \ 3.42}$

              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: