- Initial velocity of the jet, \(v_i = 153 \text{ km/h}\)
- Final velocity of the jet, \(v_f = 0 \text{ km/h}\)
- Distance traveled by the jet, \(d = 300 \text{ m}\)
- Time taken by the jet, \(t = 2.0 \text{ s}\)
To find:
- Acceleration of the jet, \(a\)
Solution:
First, we need to convert the initial velocity from km/h to m/s:
$$v_i = 153 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 42.5 \text{ m/s}$$
Now, we can use the second equation of motion to find the acceleration of the jet:
$$v_f^2 = v_i^2 + 2ad$$
Substituting the given values, we get:
$$(0 \text{ m/s})^2 = (42.5 \text{ m/s})^2 + 2a(300 \text{ m})$$
Simplifying the equation, we get:
$$a = \frac{(0 \text{ m/s})^2 - (42.5 \text{ m/s})^2}{2(300 \text{ m})}$$
$$a = \frac{-1806.25 \text{ m}^2/\text{s}^2}{600 \text{ m}}$$
$$a = -3.01 \text{ m/s}^2$$
Therefore, the acceleration of the jet is -3.01 m/s², indicating that it is decelerating at a rate of 3.01 m/s² to bring it to a stop on the aircraft carrier.
