Question

Question: A jumbo jet must reach a speed of 360 km/hron the runway for takeoff. What is the lowest constant acceleration needed for takeoff from a 1.80 km runway?

Short answer

The lowest constant acceleration of the jumbo jet is 2.78 m/s².

Step-by-step solution

Given data

Initial speed, $v_0=0\text{km/h}$

Final speed, role="math" localid="1650541085444" $v=360\text{km/h}$

Distance,$x=1.80\text{km}$

Understanding the kinematic equations Kinematic equations describe the motion of an object with constant acceleration. These equations can be used to determine the acceleration.

The expression for the kinematic equations of motion are given as follows:

$v^2={v_0}^2+2ax$ … (i)

Here, $v_0$is the initial velocity, $v$is the final velocity, $a$is the acceleration and $x$is the displacement.

Determination of the lowest constant acceleration.

Convert the velocity from km/h to m/s as:

$$\begin{gathered} v=360\frac{\text{km}}{\text{h}}\times \frac{1000\text{m}}{1\text{km}}\times \frac{1\text{h}}{3600\text{s}} \\ =100\text{m/s} \end{gathered}$$

Using equation (i), the acceleration is calculated as follows:

role="math" localid="1650541810726" $$\begin{gathered} a=\frac{v^2-{v_0}^2}{2x} \\ =\frac{{\left(100\text{m/s}\right)}^2-{\left(0\text{m/s}\right)}^2}{2\times \left(1.8\text{km×}{\displaystyle \frac{1000\text{m}}{1\text{km}}}\right)} \\ =2.78{\text{m/s}}^2 \end{gathered}$$

Thus, the lowest constant acceleration needed for the jumbo jet is $2.78{\text{m/s}}^2$.