Question

A light plane must reach a speed of 35 m/s for takeoff. What length of a runway is needed if the (constant) acceleration is 3.0 m/s²?

Short answer

A runway of length 204.2 m is needed.

Step-by-step solution

Step 1. Relation between initial velocity, final velocity, acceleration, and displacement

The rate of change in the object’s displacement is known as the object’s velocity. It has both direction and magnitude.

Given data.

The initial velocity is $v_o = 0$as it starts from rest.

The final velocity is $v_{} = 35 \frac{\mathrm{m}}{\mathrm{s}}$.

The acceleration of the plane is $a = 3.0 \frac{\mathrm{m}}{{\mathrm{s}}^2}$.

Assumption.

Assume that a runway of length $\Delta x$is needed.

Now, you know that

role="math" localid="1642833471252" $v^2 = v_o^2 + 2a \Delta x\dots \left(i\right)$.

Step 2. Calculation of the length of the highway

Now, from equation (i),

$\begin{array}{c}{v}^2 = v_o^2 + 2a \Delta x\\ {\left(35 \frac{\mathrm{m}}{\mathrm{s}}\right)}^2 = {\left(0 \frac{\mathrm{m}}{\mathrm{s}}\right)}^2 + 2 \times 3.0 \frac{\mathrm{m}}{{\mathrm{s}}^2} \times \Delta x\\ 6.0 \frac{\mathrm{m}}{{\mathrm{s}}^2} \times \Delta x = {\left(35 \frac{\mathrm{m}}{\mathrm{s}}\right)}^2 \\ \Delta x = 204.2 \mathrm{m}\end{array}$

Therefore, a runway of length 204.2 m is needed.