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 9.80 m/s in 3.10 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 jogger accelerates from rest to 5.64 m/s in 3.36 s. A car accelerates from 24.6 to 39.8 m/s also in 3.36 s. (a) Find the magnitude of the acceleration of the jogger. (b) Determine the magnitude of the acceleration of the car. (c) How much further does the car travel than the jogger during the 3.36 s?
To solve this problem, we can use the equations of motion. Let's break it down step by step:
(a) To find the bird's acceleration, we can use the equation:
acceleration = (change in velocity) / (time)
Given:
Initial velocity (u) = 13.0 m/s
Final velocity (v) = 9.80 m/s
Time (t) = 3.10 s
First, let's calculate the change in velocity:
change in velocity = v - u
change in velocity = 9.80 m/s - 13.0 m/s
change in velocity = -3.20 m/s
Now, we can substitute the values into the acceleration formula:
acceleration = (-3.20 m/s) / (3.10 s)
acceleration = -1.03 m/s^2
Therefore, the magnitude of the bird's acceleration is 1.03 m/s^2. Since the bird is running due north in a straight line, the direction of acceleration is also north.
(b) To find the bird's velocity after an additional 1.50 s, we can use the equation:
final velocity = initial velocity + (acceleration * time)
Given:
Initial velocity (u) = 9.80 m/s (since the bird slowed down to this speed)
Time (t) = 1.50 s
Substituting the values into the equation:
final velocity = 9.80 m/s + (-1.03 m/s^2 * 1.50 s)
final velocity = 9.80 m/s - 1.54 m/s
final velocity = 8.26 m/s
Therefore, assuming that the acceleration remains the same, the bird's velocity after an additional 1.50 s is 8.26 m/s.