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Complete Question:
a. A hawk flies in a horizontal arc of radius 11.3 m at a constant speed of 5.7 m/s. Find its centripetal acceleration.
Answer in units of m/s2
b. It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s2. Find the magnitude of acceleration under
these new conditions.
Answer in units of m/s2
Answer:
a. 2.875m/s²
b. 3.172m/s²
Explanation:
a. The formula for centripetal acceleration = (speed²) ÷ radius
Centripetal acceleration = (5.7m/s)²÷ 11.3m
Centripetal acceleration = 2.875m/s²
b. Magnitude of acceleration can be calculated by finding the sum of the vectors for the both the centripetal acceleration and the increase in the speed rate.
Centripetal acceleration ( acceleration x) = 2.875m/s²
Increase in the speed rate ( acceleration n) = 1.34m/s²
Magnitude of acceleration = √a²ₓ + a²ₙ
=√( 2.875m/s²)²+ (1.34m/s²)²
= √ 10.06m/s²
= 3.172m/s²
Let us assume that rocket only runs in initial energy and not using its own to flying.
Also , let upward direction is +ve and downward direction is -ve .
Initial velocity , u = 58.8 m/s .
Acceleration due to gravity ,
.
Final velocity , v - = 0 m/s .
We know , by equation of motion .

Hence, this is the required solution .
It is called Arrhenius collision theory.
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