A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 meters a second at the end of the catapult. Assuming the acceleration is constant, how far did it travel during those 2 seconds?
To find the distance traveled by the jet plane during those 2 seconds, we can use the formula:
distance = initial velocity * time + (1/2) * acceleration * time^2
We are given:
initial velocity (u) = 0 (since the jet plane starts from rest)
final velocity (v) = 42 m/s
time (t) = 2.0 s
Since the acceleration is constant, we can use the formula to find the acceleration (a):
v = u + a * t
42 = 0 + a * 2.0
Rearranging the equation, we get:
a = 42 / 2.0
a = 21 m/s^2
Now we can substitute the values into the distance formula:
distance = 0 * 2.0 + (1/2) * 21 * (2.0)^2
distance = 0 + (1/2) * 21 * 4
distance = 0 + 42
distance = 42 meters
Therefore, the jet plane traveled a distance of 42 meters during those 2 seconds.
To find out how far the jet plane traveled during those 2 seconds, we can use the equation of acceleration:
\(v = u + at\)
where:
- \(v\) is the final velocity
- \(u\) is the initial velocity
- \(a\) is the acceleration
- \(t\) is the time
In this case, the final velocity (\(v\)) is given as 42 m/s, the initial velocity (\(u\)) is 0 m/s (since the jet starts from rest), and the time (\(t\)) is 2.0 s.
We need to find the acceleration (\(a\)) in order to solve for the distance traveled. The formula for acceleration is:
\(a = \frac{{v - u}}{{t}}\)
Substituting the given values, we get:
\(a = \frac{{42 - 0}}{{2.0}}\)
\(a = 21 \, \text{m/s}^2\)
Now, to find the distance (\(s\)) traveled during those 2 seconds, we can use the equation of motion:
\(s = ut + \frac{1}{2}at^2\)
Substituting the known values, we get:
\(s = 0 \cdot 2.0 + \frac{1}{2} \cdot 21 \cdot (2.0)^2\)
\(s = 0 + \frac{1}{2} \cdot 21 \cdot 4\)
\(s = 0 + 42\)
\(s = 42 \, \text{m}\)
Therefore, the jet plane traveled a distance of 42 meters during those 2 seconds.
a = 42m/s / 2s = 21m/s^2
s = 1/2 at^2