Question

airplane starting from rest takes 25 second and 500 meter of run way at constant acceleration to leave the ground. its velocity when it becomes airborne is

Answer 1

To find the velocity of the airplane when it becomes airborne, we can use the equations of motion that relate distance, initial velocity, time, and acceleration. Since the airplane starts from rest, the initial velocity (\(u\)) is 0 m/s.

We can use the equation:

\[
s = ut + \frac{1}{2} a t^2
\]

Where:
- \(s\) is the distance (500 m)
- \(u\) is the initial velocity (0 m/s)
- \(a\) is the acceleration
- \(t\) is the time (25 s)

Substituting the known values into the equation:

\[
500 = 0 \cdot 25 + \frac{1}{2} a (25^2)
\]

This simplifies to:

\[
500 = \frac{1}{2} a (625)
\]

Multiplying both sides by 2:

\[
1000 = 625a
\]

Now, we can solve for \(a\):

\[
a = \frac{1000}{625} = 1.6 \, \text{m/s}^2
\]

Next, we will find the final velocity (\(v\)) when the airplane becomes airborne using the equation:

\[
v = u + at
\]

Substituting the values we have:

\[
v = 0 + (1.6 \cdot 25)
\]

Calculating this:

\[
v = 1.6 \cdot 25 = 40 \, \text{m/s}
\]

Thus, the velocity of the airplane when it becomes airborne is **40 m/s**.