Question

A ladybug with a velocity of 10.0 mm/s [W] crawls on a chair that is being pulled [W 50° N] at 40.0 mm/s. What is the velocity of the ladybug relative to the ground?

Answer 1

Question #53208, Physics / Mechanics | Kinematics | Dynamics

A ladybug with a velocity of $10.0 \, \text{mm/s}$ [W] crawls on a chair that is being pulled $[W \, 50{}^{\circ} \, N]$ at 40.0 mm/s. What is the velocity of the ladybug relative to the ground?

Answer:

The figure for the velocities can be drawn as follows:

Depicted parameters are:

$v_{1}$ is the velocity of the ladybug and $v_{2}$ is the velocity of chair.

The projection of $\mathbf{v}_2$ to $W$ direction is defined by the equation:

$$v _ {2} (W) = \cos 5 0 {}^ {\circ} \times v _ {2}$$

Thus, for the West direction, the velocity of the ladybug relative to the ground equals:

$$v (W) = v _ {1} + v _ {2} (W) = v _ {1} + \cos 5 0 {}^ {\circ} \times v _ {2},$$ $$v (W) = 1 0 \mathrm {m m} / \mathrm {s} + 0. 6 4 2 7 8 8 \times 4 0 \mathrm {m m} / \mathrm {s} = 3 5. 7 1 \mathrm {m m} / \mathrm {s}$$

For the North direction is defined: $\mathbf{v}(\mathbf{N}) = \mathbf{v}_1(\mathbf{N}) + \mathbf{v}_2(\mathbf{N})$ .

Taking into account that $\mathsf{v}_1(\mathsf{N}) = 0, \mathsf{v}_2(\mathsf{N}) = \sin 50{}^\circ \times \mathsf{v}_2$ , the velocity in the North direction equals:

$$v (N) = v _ {1} (N) + \sin 5 0 {}^ {\circ} \times v _ {2} = 0 + 0. 7 6 6 \times 4 0 m m / s = 3 0. 6 4 m m / s.$$

The total the velocity of the ladybug relative to the ground is determined by the equation:

$$v ^ {2} = v (N) ^ {2} + v (W) ^ {2}$$

Thus, $\mathrm{v} = \sqrt{\mathrm{v}(\mathrm{N})2 + \mathrm{v}(\mathrm{W})2} = \sqrt{1275.2041 + 938.8096} = 47.053 \, \mathrm{mm/s}$

http://www.AssignmentExpert.com/