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 11.0 m/s in 2.40 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 1.50 s has elapsed?
(a) a = (11-13 m/s)/2.4s = -2/2.4 m/s^2
(b) v = 11 - at, so use t=1.5
To find the answers to these questions, we need to use the formulas of acceleration and velocity.
(a) The magnitude of acceleration can be calculated using the formula:
acceleration = (final velocity - initial velocity) / time
In this case, the initial velocity is 13.0 m/s, the final velocity is 11.0 m/s, and the time is 2.40 s. Plugging these values into the formula, we get:
acceleration = (11.0 m/s - 13.0 m/s) / 2.40 s
= -2.0 m/s / 2.40 s
= -0.83 m/s^2
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity (north).
Therefore, the magnitude of the bird's acceleration is 0.83 m/s^2, and the direction is south.
(b) To find the bird's velocity after an additional 1.50 s has elapsed, we can use the formula:
final velocity = initial velocity + acceleration * time
The initial velocity is 11.0 m/s (since we are continuing from part (a)), the acceleration is -0.83 m/s^2, and the time is 1.50 s. Plugging these values into the formula, we get:
final velocity = 11.0 m/s + (-0.83 m/s^2) * 1.50 s
= 11.0 m/s - 1.245 m/s
= 9.75 m/s
Therefore, the bird's velocity after an additional 1.50 s has elapsed is 9.75 m/s, in the same direction as the initial velocity (north).