part 1 of 3: A hawk flies in a horizontal arc of radius 9.9 m at a constant speed of 4.8 m/s. Find its centripetal acceleration. Answer in units of m/s^2.
part 2 of 3: It continues to fly along the same horizontal arc but increases its speed at the rate of 1 m/s^2. Find the magnitude of acceleration under these new conditions. Answer in units of m/s^2.
Part 3 of 3: Find the direction of acceleration relative to the direction of motion under these conditions. Answer between -180 degrees and 180 degrees. Answer in unit of degrees.
Part 1 of 3:
To find the centripetal acceleration of the hawk, we can use the formula:
ac = v^2 / r
where ac is the centripetal acceleration, v is the velocity, and r is the radius of the arc.
Given:
v = 4.8 m/s
r = 9.9 m
Substituting the given values into the formula, we get:
ac = (4.8 m/s)^2 / 9.9 m
ac = 23.04 m^2/s^2 / 9.9 m
ac ≈ 2.33 m/s^2
Therefore, the centripetal acceleration of the hawk is approximately 2.33 m/s^2.
Part 2 of 3:
When the hawk increases its speed at a rate of 1 m/s^2, we need to consider both the initial speed and the rate of change of speed.
Let's denote the initial speed as v0 and the rate of change of speed as a.
Given:
v0 = 4.8 m/s
a = 1 m/s^2
r = 9.9 m
To find the magnitude of acceleration, we can use the equation:
a = √[(ac)^2 + (at)^2]
where ac is the centripetal acceleration, at is the tangential acceleration, and a is the magnitude of acceleration.
The tangential acceleration can be calculated using the equation:
at = dv / dt
where dv is the change in velocity and dt is the change in time.
Since the hawk is changing its speed at a constant rate, we can consider dt to be equal to 1 second (since it is given in the unit of seconds).
Substituting the given values into the formula, we get:
at = (1 m/s^2) * (1 s)
at = 1 m/s
Now, we can substitute the values for ac and at into the equation for magnitude of acceleration:
a = √[(2.33 m/s^2)^2 + (1 m/s)^2]
a ≈ √[5.4289 m^2/s^4 + 1 m^2/s^2]
a ≈ √[6.4289 m^2/s^4]
a ≈ 2.54 m/s^2
Therefore, the magnitude of acceleration under the new conditions is approximately 2.54 m/s^2.
Part 3 of 3:
To find the direction of acceleration relative to the direction of motion, we need to consider the components of acceleration in the horizontal and vertical directions.
Since the horizontal motion is in a circle, the centripetal acceleration is pointing toward the center of the circle.
The direction of acceleration can be found using the equation:
θ = arctan(at/ac)
where θ is the angle between acceleration and motion, ac is the centripetal acceleration, and at is the tangential acceleration.
Substituting the given values into the formula, we get:
θ = arctan((1 m/s) / (2.33 m/s^2))
θ ≈ arctan(0.429)
θ ≈ 23.8 degrees
Therefore, the direction of acceleration relative to the direction of motion under these conditions is approximately 23.8 degrees.
To find the centripetal acceleration in part 1, we can use the formula:
Centripetal acceleration (a) = (v^2) / r
where v is the speed of the hawk and r is the radius of its arc.
Given that the speed of the hawk (v) is 4.8 m/s and the radius (r) is 9.9 m, we can substitute these values into the formula:
a = (4.8^2) / 9.9
Calculating this expression gives us:
a ≈ 2.323 m/s²
So, the magnitude of the centripetal acceleration of the hawk is approximately 2.323 m/s².
Moving on to part 2, we need to find the magnitude of acceleration when the speed of the hawk increases at a rate of 1 m/s².
The magnitude of acceleration, in this case, can be calculated using the formula:
Acceleration (a) = Centripetal acceleration (a_c) + Tangential acceleration (a_t)
Since the speed is changing, there is a tangential component of acceleration caused by the change in speed. The centripetal acceleration remains the same as before.
From part 1, we know that the centripetal acceleration is approximately 2.323 m/s².
To calculate the tangential acceleration (a_t), we can use the following formula:
a_t = dv / dt
Given that the rate of change of speed is 1 m/s², we can substitute this value into the formula:
a_t = 1 m/s²
Therefore, the magnitude of acceleration (a) in part 2 is the sum of the centripetal acceleration (a_c) and tangential acceleration (a_t):
a = a_c + a_t
= 2.323 m/s² + 1 m/s²
= 3.323 m/s²
So, the magnitude of acceleration under the new conditions is 3.323 m/s².
Finally, in part 3, we need to find the direction of acceleration relative to the direction of motion under these new conditions.
Since the direction of acceleration is always directed towards the center of the circular path, while the direction of motion is tangential to the circle, the two directions are always perpendicular to each other.
Therefore, the direction of acceleration relative to the direction of motion is 90 degrees or -90 degrees.
In summary:
Part 1: The centripetal acceleration is approximately 2.323 m/s².
Part 2: The magnitude of acceleration under the new conditions is 3.323 m/s².
Part 3: The direction of acceleration relative to the direction of motion is 90 degrees or -90 degrees.