An airplane travels 330 m down the runway before taking off. Assuming that it has constant acceleration, if it starts from rest and becomes airborne in 9.00 s, how fast (in m/s) is it moving at takeoff?
A 73.3 m/s
B 36.7 m/s
C 1490 m/s
D 660 m/s
s = 1/2 at^2, so a = 2s/t^2
v = at = (2s/t^2)t = 2s/t = 73.3 m/s
To find the speed of the airplane at takeoff, we can use the formula for constant acceleration motion:
v = u + at
where:
v = final velocity (speed at takeoff)
u = initial velocity (speed at rest, which is 0 m/s)
a = acceleration
t = time taken to become airborne (9.00 s)
Since the airplane starts from rest (u = 0 m/s), the equation simplifies to:
v = at
To find the acceleration, we need the distance traveled. The distance traveled is given as 330 m.
The formula to find acceleration is:
a = (v - u) / t
Since the airplane starts from rest (u = 0 m/s), the equation simplifies to:
a = v / t
Rearranging the equation to solve for v:
v = a * t
Substituting the given values:
v = (330 m) / (9.00 s)
v ā 36.7 m/s
Therefore, the airplane is moving at a speed of approximately 36.7 m/s at takeoff.
Answer: B) 36.7 m/s
To find the speed of the airplane at takeoff, we can use the kinematic equation:
v = u + at
Where:
- v is the final velocity (speed) of the airplane at takeoff,
- u is the initial velocity (speed) of the airplane at rest,
- a is the acceleration of the airplane, and
- t is the time taken for the airplane to become airborne.
In this case, the airplane starts from rest, so its initial velocity is 0 m/s (u = 0). We're given that the airplane becomes airborne in 9.00 s (t = 9.00 s). We need to find the acceleration (a) to determine the final velocity (v).
To find the acceleration, we can use the equation:
d = ut + (1/2)at^2
Where:
- d is the distance traveled by the airplane down the runway before takeoff.
The distance traveled down the runway before takeoff is given as 330 m (d = 330 m). Since the airplane starts from rest, its initial velocity is 0 m/s (u = 0). We can simplify the equation to solve for a:
330 m = (1/2)a(9.00 s)^2
330 m = (1/2)a(81 s^2)
660 m = 81a
Now we can solve for a:
a = 660 m / 81 s^2
a ā 8.148 m/s^2
Now that we have the acceleration, we can substitute it back into the first kinematic equation to find the final velocity:
v = u + at
v = 0 m/s + (8.148 m/s^2)(9.00 s)
v ā 73.333 m/s
So, the airplane is moving at approximately 73.3 m/s at takeoff.
Therefore, the correct answer is A) 73.3 m/s.