1)A hawk flies in a horizontal arc of radius 16 m at a constant speed of 4.8 m/s. Find its centripetal acceleration. Answer in units of m/s2.
2)It continues to fly along the same horizontal arc but increases its speed at the rate of 1.52 m/s2. Find the magnitude of acceleration under these new conditions. Answer in units of m/s2.
3)Find the direction of acceleration relative to the direction of motion under these conditions. Answer between −180◦ and 180◦. Answer in units of ◦.
centripetal acceleration, directed towards the center, is v^2/r
2) the tangential acceleration is 1.52m/s^2, directed at 90 degrees to the center. Add the two accelerations as vectors.
1) To find the centripetal acceleration of the hawk, we can use the formula:
\(a_c = \frac{v^2}{r}\)
where \(a_c\) is the centripetal acceleration, \(v\) is the speed of the hawk, and \(r\) is the radius of the arc.
Given that the hawk's speed is 4.8 m/s and the radius is 16 m, we can substitute these values into the formula:
\(a_c = \frac{(4.8)^2}{16}\)
Simplifying the equation:
\(a_c = \frac{23.04}{16}\)
\(a_c = 1.44\) m/s²
Therefore, the centripetal acceleration of the hawk is 1.44 m/s².
2) The magnitude of acceleration under the new conditions can be found by considering both the constant centripetal acceleration and the additional acceleration due to the increase in speed.
The centripetal acceleration remains the same as before, which is 1.44 m/s².
The additional acceleration due to the increase in speed can be determined using the formula:
\(a_{additional} = \frac{dv}{dt}\)
where \(a_{additional}\) is the additional acceleration, \(dv\) is the change in velocity, and \(dt\) is the change in time.
Given that the rate of increase in speed is 1.52 m/s², we can substitute this value into the formula:
\(a_{additional} = 1.52\) m/s²
Therefore, the magnitude of acceleration under the new conditions is the sum of the centripetal acceleration (1.44 m/s²) and the additional acceleration (1.52 m/s²):
\(a_{total} = 1.44 + 1.52\)
\(a_{total} = 2.96\) m/s²
So, the magnitude of acceleration under these new conditions is 2.96 m/s².
3) To determine the direction of acceleration relative to the direction of motion, we can look at whether the acceleration is towards the center of the circle (centripetal acceleration) or tangential to the motion (tangential acceleration).
In this case, since the hawk is increasing its speed, the acceleration is both centripetal and tangential. The centripetal acceleration is always directed towards the center of the circle, while the tangential acceleration is directed tangentially to the motion.
Since the hawk is moving in a horizontal arc, the tangential acceleration is in the horizontal direction, which can be considered as the positive direction.
So, the direction of acceleration relative to the direction of motion is 0 degrees (or simply positive direction).
Therefore, the direction of acceleration relative to the direction of motion under these conditions is 0°.