The question is not complete and the complete question is ;
"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
part 2
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:
1) Centripetal Accelaration = 2.875 m/s²
2) Magnitude of acceleration is 3.172 m/s² and direction is at an angle of 65º from tangent toward the center of the arc.
Explanation:
Centripetal force = mv²/r
Where m = mass, v=velocity and r = Radius
Also,
Force = ma
Where a is centripetal accelaration
Thus,
mv²/r = ma
a = v²/r
a = (5.7² ÷ 11.3) = 2.875 m/s²
The direction of the velocity of an object moving around an arc is tangent to the arc and thus this velocity is called tangential velocity!
From the question, it increases its tangential velocity at the rate of 1.34 m/s², the hawk must exert a force in the same direction as the velocity.
This force will cause a tangential acceleration in the same direction as the tangential velocity.
The direction of the centripetal acceleration is along a radial line pointing toward the center of the arc while the direction of the tangential acceleration is perpendicular to the radial line pointing toward the center of the arc.
Thus, the 2.875 m/s² centripetal acceleration is perpendicular to the 1.34 m/s² tangential acceleration.
So, magnitude =√(2.875² + 1.34²) = √10.061225 = 3.172 m/s²
Tangent of angle = 2.875 ÷ 1.34 = 2.1455
Angle = tan^-1 (2.1455)= 65º
Thus,the hawk is accelerating at 3.172 m/s² at an angle of 65º from tangent toward the center of the arc.
