A small private airplane, starting from rest on a Gander airport runway, accelerates for 19.0 s before taking off. Its speed at takeoff is 56.0 m/s (125 mi/hr).
I don't mind helping to find answers, but I resent having to provide the questions as well.
To find the acceleration of the small private airplane, we can use the formula:
\[ v = u + at \]
Where:
- v is the final velocity (speed at takeoff)
- u is the initial velocity (starting from rest)
- a is the acceleration of the airplane
- t is the time it takes for the acceleration to occur
Given:
v = 56.0 m/s
u = 0 (starting from rest)
t = 19.0 s
We can rearrange the formula to solve for acceleration (a):
\[ a = \frac{{v - u}}{{t}} \]
Substituting the given values:
\[ a = \frac{{56.0 \, \text{m/s} - 0}}{{19.0 \, \text{s}}} \]
Simplifying the equation:
\[ a = \frac{{56.0 \, \text{m/s}}}{{19.0 \, \text{s}}} \]
Therefore, the acceleration of the small private airplane is:
\[ a = 2.95 \, \text{m/s}^2 \]
To solve this problem, we need to find the acceleration of the airplane during the 19.0 s interval.
Step 1: Convert the speed at takeoff from miles per hour to meters per second.
Given: speed at takeoff = 56.0 m/s
Step 2: Find the acceleration.
Given: time = 19.0 s, initial velocity (starting from rest) = 0 m/s, final velocity (at takeoff) = 56.0 m/s
The formula to calculate acceleration is:
acceleration = (final velocity - initial velocity) / time
acceleration = (56.0 m/s - 0 m/s) / 19.0 s
Step 3: Calculate the acceleration.
acceleration = 56.0 m/s / 19.0 s
acceleration ≈ 2.947 m/s² (rounded to 3 decimal places)
Therefore, the acceleration of the airplane during the 19.0 s interval is approximately 2.947 m/s².