An airplane taxis onto the runway going at 10 m/s.if it can accelerate at 3.0 m/s square and its take-off speed is 90 m/s, what length of runway will it need?

vf^2=vi^2+2ad solve for d

150m

1.5km

To find the length of runway the airplane will need, we can use the equations of motion. In this case, we'll use the following equation:

\[v^2 = u^2 + 2as\]

Where:
- \(v\) is the final velocity (take-off speed) of the airplane,
- \(u\) is the initial velocity (taxi speed) of the airplane,
- \(a\) is the acceleration of the airplane,
- \(s\) is the distance (length of the runway) the airplane will need.

Let's plug in the given values into the equation:

\(v = 90 \, \text{m/s}\) (take-off speed)
\(u = 10 \, \text{m/s}\) (initial speed)
\(a = 3.0 \, \text{m/s}^2\) (acceleration)

Now, rearranging the equation to solve for \(s\):

\[s = \frac{{v^2 - u^2}}{{2a}}\]

Substituting the given values:

\[s = \frac{{(90 \, \text{m/s})^2 - (10 \, \text{m/s})^2}}{{2 \times 3.0 \, \text{m/s}^2}}\]

Now, let's calculate the value of \(s\):

\[s = \frac{{8100 \, \text{m}^2/\text{s}^2 - 100 \, \text{m}^2/\text{s}^2}}{{6.0 \, \text{m/s}^2}}\]

\[s = \frac{{8000 \, \text{m}^2/\text{s}^2}}{{6.0 \, \text{m/s}^2}}\]

\[s = \frac{{8000}}{{6.0}} \, \text{m}\]

\[s \approx 1333.33 \, \text{m}\]

Therefore, the airplane will need a runway length of approximately 1333.33 meters.