A plane starting at rest at the south end of a runway undergoes a uniform acceleration of 1.60 m/s^2 to the north. At takeoff, the plane's velocity s 72.0 m/s to the north.
A. What is the time required for takeoff?
B. How far does the plane travel along the runway?
To find the time required for takeoff, we can use the following kinematic equation:
v = u + at
where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time
For part A, we are given:
- u (initial velocity) = 0 m/s (plane is at rest)
- v (final velocity) = 72.0 m/s
- a (acceleration) = 1.60 m/s^2
Substituting the known values into the equation:
72.0 m/s = 0 m/s + 1.60 m/s^2 * t
Simplifying the equation:
72.0 m/s = 1.60 m/s^2 * t
Now we can solve for t by rearranging the equation:
t = (72.0 m/s) / (1.60 m/s^2)
Calculating the value:
t = 45 seconds
Therefore, the time required for takeoff is 45 seconds.
For part B, we can use the equation of motion:
s = ut + (1/2)at^2
where:
- s is the distance traveled
- u is the initial velocity
- a is the acceleration
- t is the time
Since the plane starts at rest (u = 0), the equation simplifies to:
s = (1/2)at^2
Substituting the known values:
s = (1/2) * (1.60 m/s^2) * (45 s)^2
Calculating the value:
s = 1620 meters
Therefore, the plane travels a distance of 1620 meters along the runway.