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 for centripetal acceleration, which is given by the equation a = v^2 / r, where v represents the velocity of the object in meters per second (m/s) and r represents the radius of the circular path in meters (m).
In this case, the velocity of the hawk is given as 4.8 m/s, and the radius of the arc it is flying in is 9.9 m. Plugging these values into the formula, we get:
a = (4.8 m/s)^2 / 9.9 m
Simplifying the equation, we have:
a = 23.04 m^2/s^2 / 9.9 m
Dividing the values, we find:
a ≈ 2.327 m/s^2
Therefore, the centripetal acceleration of the hawk is approximately 2.327 m/s^2.
Part 2 of 3: Under the new conditions where the hawk increases its speed at a rate of 1 m/s^2, we need to find the magnitude of the acceleration. In this case, we can use a similar formula for calculating the magnitude of acceleration, which is given by the equation a = √(a_c^2 + a_t^2), where a_c represents the centripetal acceleration and a_t represents the tangential acceleration.
In this scenario, the tangential acceleration is simply the given rate at which the speed is increasing, which is 1 m/s^2. The centripetal acceleration remains the same as calculated in Part 1, which was approximately 2.327 m/s^2.
Using the formula, we find:
a = √((2.327 m/s^2)^2 + (1 m/s^2)^2)
Simplifying the equation, we have:
a = √( 5.4083 m^2/s^4 + 1 m^2/s^4)
Taking the square root, we get:
a ≈ √6.4083 m^2/s^4
Therefore, the magnitude of the acceleration under these new conditions is approximately 2.529 m/s^2.
Part 3 of 3: To find the direction of the acceleration relative to the direction of motion, we can use the concept of vectors. The direction of centripetal acceleration is always towards the center of the circular path, while the direction of tangential acceleration is in the direction of the velocity.
In this case, the acceleration is made up of two components: centripetal acceleration and tangential acceleration. The centripetal acceleration always points towards the center of the circular path, while the tangential acceleration points along the direction of motion.
Since the tangential acceleration is positive in this case (due to an increase in speed), and the centripetal acceleration is constant and directed towards the center of the circle, the direction of the resulting acceleration will be a combination of the two.
To determine the direction, we can use the trigonometric relationship between the two components. The angle between the centripetal acceleration and the resulting acceleration can be found using the inverse tangent function, given by the equation θ = tan^(-1)(a_t / a_c).
Using the values we calculated in Part 1 (a_c ≈ 2.327 m/s^2) and Part 2 (a_t = 1 m/s^2), we find:
θ = tan^(-1)(1 m/s^2 / 2.327 m/s^2)
Calculating the inverse tangent, we get:
θ ≈ 24.38 degrees
Therefore, the direction of acceleration relative to the direction of motion under these conditions is approximately 24.38 degrees.