Chapter 3, Problem 50E

(a)

To determine

The bird’s speed relative to the ground.

(b)

To determine: The acceleration of the bird (Magnitude and direction)

Answer: The acceleration of the bird is $9.457{\text{m/s}}^{\text{2}}$ radial outward to the circle.

Explanation:

When a body moves in the circular path then it experience an acceleration known as centrifugal acceleration in radial outward direction of the circle.

$a=\frac{v^2}{r}$

• $a$ is the acceleration

Substitute $7.54\text{m/s}$ for v and 6 m for r in the above equation.

$\begin{array}{c}a=\frac{{\left(7.54\text{m/s}\right)}^2}{\left(6\text{m}\right)}\\ =9.457{\text{m/s}}^{\text{2}}\end{array}$

Thus, the acceleration of the bird is $9.457{\text{m/s}}^{\text{2}}$ radial outward to the circle.

Conclusion:

Hence, acceleration of the bird is $\boxed{9.457{\text{m/s}}^{\text{2}}}$.

(c)

To determine

Angle between the bird velocity and ground.