Question

An Australian emu is running due north in a straight line at a speed of 13.0 m/s and slows down to a speed of 9.40 m/s in 3.50 s. (a) What is the magnitude and direction of the bird’s acceleration? (b) Assuming that the acceleration remains the same, what is the bird’s velocity after an additional 1.20 s has elapsed?

Answer 1

Answer:

a) Magnitude = 1.03 m/s², Direction: south

b) $V_{f} = 8.16 m/s$

Explanation:

a) The magnitude and direction of the acceleration can be calculated using the following equation:

$V_{f} = V_{0} + at$     (1)

Where:

$V_{f}$: is the final speed = 9.40 m/s  

$V_{0}$: is the initial speed = 13.0 m/s

t: is the time = 3.50 s

Solving equation (1) for a, we have:

$a = \frac{V_{f} - V_{0}}{t} = \frac{9.40 m/s - 13.0 m/s}{3.50 s} = -1.03 m/s^{2}$

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.  

b) The final velocity of the bird can be found using the same equation 1:

$V_{f} = V_{0} + at$

$V_{f} = 13.0 m/s + (-1.03 m/s^{2})*(3.50 s + 1.20 s) = 8.16 m/s$

Therefore, the bird’s velocity after an additional 1.20 s has elapsed is 8.16 m/s.

I hope it helps you!