A CF-18 fighter jet flying at 350 m/ s engages its afterburners and accelerates at a rate of 12.6 m/ s 2 to a velocity of 600 m/ s. How far does the fighter jet travel during acceleration?
To find the distance traveled during acceleration, we can use the equation:
\[v^2 = u^2 + 2as\]
Where:
- v is the final velocity (600 m/s)
- u is the initial velocity (350 m/s)
- a is the acceleration (12.6 m/s^2)
- s is the distance traveled
Rearranging the equation, we have:
\[s = \frac{v^2 - u^2}{2a}\]
Plugging in the values, we get:
\[s = \frac{(600^2) - (350^2)}{2(12.6)}\]
Simplifying the equation, we get:
\[s = \frac{360000 - 122500}{25.2}\]
\[s = \frac{237500}{25.2}\]
\[s \approx 9434.92 \text{ meters}\]
Therefore, the fighter jet travels approximately 9434.92 meters (or 9.43 kilometers) during acceleration.
A butterfly accelerates over a distance of 10 cm in 3.0 s, increasing its velocity to 5.0 cm/ s. What was its initial velocity?
To find the initial velocity of the butterfly, we can use the equation:
\[v = u + at\]
Where:
- v is the final velocity (5.0 cm/s)
- u is the initial velocity (unknown)
- a is the acceleration (unknown)
- t is the time (3.0 s)
We also know that the distance traveled is given by the equation:
\[s = ut + \frac{1}{2}at^2\]
Substituting the values, we get:
\[0.10 m = u(3.0s) + \frac{1}{2}a(3.0s)^2\]
Simplifying the equation, we have:
\[0.10 m = 3.0u + (4.5)a\]
This is the first equation.
We also know that the final velocity is given by:
\[v^2 = u^2 + 2as\]
\[25 cm^2/s^2 = u^2 + 2a(0.10 m)\]
Converting the distance to centimeters:
\[25 cm^2/s^2 = u^2 + 20a\]
This is the second equation.
We now have a system of two equations. Let's solve it.
First, we rearrange the first equation to isolate a variable. Solving for a, we have:
\[a = \frac{-3.0u + 0.10}{4.5}\]
Substituting this expression for a into the second equation, we get:
\[25 cm^2/s^2 = u^2 + 20\left(\frac{-3.0u + 0.10}{4.5}\right)\]
Expanding and simplifying, we have:
\[25 = u^2 - 4u + 0.04\]
Rearranging the equation, we get:
\[u^2 - 4u - 24.96 = 0\]
Now we can solve for u by using the quadratic formula:
\[u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In our case, a = 1, b = -4, and c = -24.96. Plugging these values in, we get:
\[u = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-24.96)}}{2(1)}\]
Simplifying, we have:
\[u = \frac{4 \pm \sqrt{16 + 99.84}}{2}\]
\[u = \frac{4 \pm \sqrt{115.84}}{2}\]
\[u = \frac{4 \pm 10.764}{2}\]
We have two possible solutions for u:
1) \[u = \frac{4 + 10.764}{2} = 7.882 \text{ cm/s}\]
2) \[u = \frac{4 - 10.764}{2} = -3.882 \text{ cm/s}\]
Since velocity cannot be negative in this context, the initial velocity of the butterfly is approximately 7.882 cm/s.
To calculate the distance the fighter jet travels during acceleration, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (600 m/s)
u = initial velocity (350 m/s)
a = acceleration (12.6 m/s^2)
s = distance
Rearranging the equation to solve for s, we have:
s = (v^2 - u^2) / (2a)
Substituting the given values into the equation:
s = (600^2 - 350^2) / (2*12.6)
s = (360000 - 122500) / 25.2
s = 237500 / 25.2
s ≈ 9434.5 meters
Therefore, the fighter jet travels approximately 9434.5 meters during acceleration.