An airplane starts from rest at the end of a runway and
accelerates at a constant rate. In the first second, the airplane
travels 1.11 m. What is the speed of the airplane at the end of the
second second?
Well, it sounds like this airplane is in a bit of a hurry! Let's calculate its speed using a little mathematical magic, shall we?
Since the airplane is accelerating at a constant rate, we can use the equation for motion with constant acceleration:
s = ut + (1/2)at^2
Here, "s" represents the distance traveled, "u" is the initial velocity (which is 0 m/s because the airplane starts from rest), "a" is the acceleration, and "t" is the time.
Given that the airplane travels 1.11 m in the first second, we can plug in these values:
1.11 = 0 + (1/2)a(1^2)
Simplifying, we find:
1.11 = (1/2)a
Now, solve for "a":
a = 1.11 x 2
a = 2.22 m/s^2
Fantastic! Now that we know the acceleration, we can find the speed of the airplane at the end of the second second. Let's use the equation:
v = u + at
Since the airplane starts from rest (u = 0), we have:
v = 0 + (2.22)(2)
v = 4.44 m/s
Voila! The airplane is traveling at a speed of 4.44 m/s at the end of the second second. It's zooming through the sky like a caffeinated seagull!
To find the speed of the airplane at the end of the second second, we need to know the acceleration of the airplane. Unfortunately, the information provided does not include the acceleration. Could you please provide the acceleration of the airplane?
To find the speed of the airplane at the end of the second second, we need to use the equations of motion.
First, let's identify the given information:
- Initial velocity (u) = 0 (since the airplane starts from rest)
- Distance traveled (s) in the first second = 1.11 m
- Time (t) = 1 second
- Acceleration (a) = constant rate of acceleration (not given)
Now, we can use the equation of motion to find the acceleration:
s = ut + (1/2)at^2
Given:
s = 1.11 m
u = 0 m/s
t = 1 s
Plugging in the values, the equation becomes:
1.11 = (0)(1) + (1/2)a(1)^2
Simplifying, we get:
1.11 = (1/2)a
To find the value of acceleration (a), we can rearrange the equation:
a = (2 * 1.11) / 1
So, the acceleration is a = 2.22 m/s^2.
Now, we can use this acceleration value to calculate the speed at the end of the second second. We can use another equation of motion:
v = u + at
Given:
u = 0 m/s
a = 2.22 m/s^2
t = 2 s (since we want to find the speed at the end of the second second)
Plugging in the values, the equation becomes:
v = 0 + (2.22)(2)
Simplifying, we get:
v = 0 + 4.44
Therefore, the speed of the airplane at the end of the second second is v = 4.44 m/s.
0.5a*1^2 = 1.11 m.
0.5a = 1.11,
a = 2.22 m/s^2,
V1 = at = 2.22 * 1 = 2.22 m/s. = Velocity after 1 sec.
V2 = Vo + at,
V2 = 2.22 + 2.22*1 = 4.44 m/s. = Velocity after 2 sec.