Respuesta :
To solve this problem, we can use the following kinematic equation:
\[ \text{Final velocity}^2 = \text{Initial velocity}^2 + 2 \times \text{Acceleration} \times \text{Displacement} \]
Here, the final velocity (\(v_f\)) is the velocity of the kite after 4.22 seconds, the initial velocity (\(v_i\)) is what we're trying to find, the acceleration (\(a\)) is given as -1.33 m/s² (negative because it's deceleration), and the displacement (\(d\)) is given as -5.80 m (negative because it's in the opposite direction of the initial velocity).
The equation becomes:
\[ v_f^2 = v_i^2 + 2 \times a \times d \]
First, we need to find the final velocity using the following kinematic equation:
\[ v_f = v_i + a \times t \]
where \(t\) is the time, which is 4.22 s.
\[ v_f = v_i + (-1.33 \, \text{m/s}^2) \times (4.22 \, \text{s}) \]
Now, we can use this final velocity in the first equation:
\[ (v_i + (-1.33 \, \text{m/s}^2) \times (4.22 \, \text{s}))^2 = v_i^2 + 2 \times (-1.33 \, \text{m/s}^2) \times (-5.80 \, \text{m}) \]
Let's solve for \(v_i\):
\[ v_i^2 + 2 \times 1.33 \times 4.22 \times v_i - (1.33 \times 4.22)^2 = v_i^2 + 2 \times 1.33 \times 5.80 \]
Simplifying:
\[ 2 \times 1.33 \times 4.22 \times v_i = 2 \times 1.33 \times 5.80 \]
\[ v_i = \frac{2 \times 1.33 \times 5.80}{2 \times 1.33 \times 4.22} \]
Now, plug in the values:
\[ v_i \approx \frac{15.308}{11.2376} \]
\[ v_i \approx 1.363 \, \text{m/s} \]
So, the initial velocity of the kite is approximately \(1.363 \, \text{m/s}\).
