Answer:
a) Magnitude = 1.03 m/s², Direction: south
b) [tex]V_{f} = 8.16 m/s [/tex]
Explanation:
a) The magnitude and direction of the acceleration can be calculated using the following equation:
[tex] V_{f} = V_{0} + at [/tex] Â Â (1)
Where:
[tex]V_{f}[/tex]: is the final speed = 9.40 m/s Â
[tex]V_{0}[/tex]: is the initial speed = 13.0 m/s
t: is the time = 3.50 s
Solving equation (1) for a, we have:
[tex] a = \frac{V_{f} - V_{0}}{t} = \frac{9.40 m/s - 13.0 m/s}{3.50 s} = -1.03 m/s^{2} [/tex]
Hence, the magnitude of the acceleration is 1.03 m/s² and the direction of the bird's acceleration is the opposite of the initial velocity direction, which means that the bird is decelerating. Â
b) The final velocity of the bird can be found using the same equation 1:
[tex] V_{f} = V_{0} + at [/tex]
[tex] V_{f} = 13.0 m/s + (-1.03 m/s^{2})*(3.50 s + 1.20 s) = 8.16 m/s [/tex]
Therefore, the bird’s velocity after an additional 1.20 s has elapsed is 8.16 m/s.
I hope it helps you!
