A radar tracks an aircraft using radial and transverse velocity components. Given radial distance (r) = 13.1000 ft; angular rate (theta_dot) = 1.1000 rad/s; radial velocity (r_dot) = 19.0000 ft/s, determine the transverse velocity (v_theta) in ft/s.

Start with the thinking

• The governing relation printed in this handbook section is Radial and transverse velocity components.
• Everything except v_theta is given, so isolate v_theta symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Radial and transverse components for planar motion describe a particle's velocity in polar coordinates.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

2. Step 2 — Rearrange symbolically for v_theta:

   $$v_{theta} = r \dot{\theta}$$

3. Step 3 — List the givens: radial distance (r) = 13.1000 ft, angular rate (theta_dot) = 1.1000 rad/s, radial velocity (r_dot) = 19.0000 ft/s.

4. Step 4 — Substitute the given values:

   $$v_{theta} = 13.1000 \dot{\theta}$$

5. Step 5 — Evaluate:

   $$v_{theta} = 14.4100\ \text{ft/s}$$

6. Step 6 — Check: returning v_theta = 14.4100 ft/s to

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 28.8200 — kept a factor of two that cancels in the correct rearrangement.
• 7.2050 — dropped that same factor in the other direction.
• 15.8510 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Radial and Transverse Components for Planar Motion

A particle sliding in a rotating slot is analyzed with radial and transverse components. Given angular rate (theta_dot) = 4.8500 rad/s; transverse velocity (v_theta) = 4.3000 ft/s; radial velocity (r_dot) = 8.1000 ft/s, determine the radial distance (r) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Radial and transverse velocity components.
• Everything except r is given, so isolate r symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Radial and transverse components for planar motion describe a particle's velocity in polar coordinates.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

2. Step 2 — Rearrange symbolically for r:

   $$r = \dfrac{v_\theta}{\dot{\theta}}$$

3. Step 3 — List the givens: angular rate (theta_dot) = 4.8500 rad/s, transverse velocity (v_theta) = 4.3000 ft/s, radial velocity (r_dot) = 8.1000 ft/s.

4. Step 4 — Substitute the given values:

   $$r = \dfrac{v_\theta}{\dot{\theta}}$$

5. Step 5 — Evaluate:

   $$r = 0.8866\ \text{ft}$$

6. Step 6 — Check: returning r = 0.8866 ft to

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 1.7732 — kept a factor of two that cancels in the correct rearrangement.
• 0.4433 — dropped that same factor in the other direction.
• 0.9753 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Radial and Transverse Components for Planar Motion

A satellite's planar motion is decomposed into radial and transverse components. Given radial distance (r) = 4.3000 ft; transverse velocity (v_theta) = 18.9000 ft/s; radial velocity (r_dot) = 11.5000 ft/s, determine the angular rate (theta_dot) in rad/s.

Start with the thinking

• The governing relation printed in this handbook section is Radial and transverse velocity components.
• Everything except theta_dot is given, so isolate theta_dot symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Radial and transverse components for planar motion describe a particle's velocity in polar coordinates.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

2. Step 2 — Rearrange symbolically for theta_dot:

   $$\theta_{dot} = \dfrac{v_\theta}{r}$$

3. Step 3 — List the givens: radial distance (r) = 4.3000 ft, transverse velocity (v_theta) = 18.9000 ft/s, radial velocity (r_dot) = 11.5000 ft/s.

4. Step 4 — Substitute the given values:

   $$\theta_{dot} = \dfrac{v_\theta}{4.3000}$$

5. Step 5 — Evaluate:

   $$\theta_{dot} = 4.3953\ \text{rad/s}$$

6. Step 6 — Check: returning theta_dot = 4.3953 rad/s to

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 8.7907 — kept a factor of two that cancels in the correct rearrangement.
• 2.1977 — dropped that same factor in the other direction.
• 4.8349 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Radial and Transverse Components for Planar Motion

A radar tracks an aircraft using radial and transverse velocity components. Given radial distance (r) = 4.0000 ft; angular rate (theta_dot) = 2.8000 rad/s; radial velocity (r_dot) = 12.7000 ft/s, determine the transverse velocity (v_theta) in ft/s.

Start with the thinking

• The governing relation printed in this handbook section is Radial and transverse velocity components.
• Everything except v_theta is given, so isolate v_theta symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Radial and transverse components for planar motion describe a particle's velocity in polar coordinates.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

2. Step 2 — Rearrange symbolically for v_theta:

   $$v_{theta} = r \dot{\theta}$$

3. Step 3 — List the givens: radial distance (r) = 4.0000 ft, angular rate (theta_dot) = 2.8000 rad/s, radial velocity (r_dot) = 12.7000 ft/s.

4. Step 4 — Substitute the given values:

   $$v_{theta} = 4.0000 \dot{\theta}$$

5. Step 5 — Evaluate:

   $$v_{theta} = 11.2000\ \text{ft/s}$$

6. Step 6 — Check: returning v_theta = 11.2000 ft/s to

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 22.4000 — kept a factor of two that cancels in the correct rearrangement.
• 5.6000 — dropped that same factor in the other direction.
• 12.3200 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Radial and Transverse Components for Planar Motion

A particle sliding in a rotating slot is analyzed with radial and transverse components. Given angular rate (theta_dot) = 1.7500 rad/s; transverse velocity (v_theta) = 52.8000 ft/s; radial velocity (r_dot) = 1.6000 ft/s, determine the radial distance (r) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Radial and transverse velocity components.
• Everything except r is given, so isolate r symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Radial and transverse components for planar motion describe a particle's velocity in polar coordinates.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

2. Step 2 — Rearrange symbolically for r:

   $$r = \dfrac{v_\theta}{\dot{\theta}}$$

3. Step 3 — List the givens: angular rate (theta_dot) = 1.7500 rad/s, transverse velocity (v_theta) = 52.8000 ft/s, radial velocity (r_dot) = 1.6000 ft/s.

4. Step 4 — Substitute the given values:

   $$r = \dfrac{v_\theta}{\dot{\theta}}$$

5. Step 5 — Evaluate:

   $$r = 30.1714\ \text{ft}$$

6. Step 6 — Check: returning r = 30.1714 ft to

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 60.3429 — kept a factor of two that cancels in the correct rearrangement.
• 15.0857 — dropped that same factor in the other direction.
• 33.1886 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Radial and Transverse Components for Planar Motion

A satellite's planar motion is decomposed into radial and transverse components. Given radial distance (r) = 8.4000 ft; transverse velocity (v_theta) = 20.7000 ft/s; radial velocity (r_dot) = 2.8000 ft/s, determine the angular rate (theta_dot) in rad/s.

Start with the thinking

• The governing relation printed in this handbook section is Radial and transverse velocity components.
• Everything except theta_dot is given, so isolate theta_dot symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Radial and transverse components for planar motion describe a particle's velocity in polar coordinates.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

2. Step 2 — Rearrange symbolically for theta_dot:

   $$\theta_{dot} = \dfrac{v_\theta}{r}$$

3. Step 3 — List the givens: radial distance (r) = 8.4000 ft, transverse velocity (v_theta) = 20.7000 ft/s, radial velocity (r_dot) = 2.8000 ft/s.

4. Step 4 — Substitute the given values:

   $$\theta_{dot} = \dfrac{v_\theta}{8.4000}$$

5. Step 5 — Evaluate:

   $$\theta_{dot} = 2.4643\ \text{rad/s}$$

6. Step 6 — Check: returning theta_dot = 2.4643 rad/s to

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 4.9286 — kept a factor of two that cancels in the correct rearrangement.
• 1.2321 — dropped that same factor in the other direction.
• 2.7107 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Radial and Transverse Components for Planar Motion

A radar tracks an aircraft using radial and transverse velocity components. Given radial distance (r) = 5.7000 ft; angular rate (theta_dot) = 2.4000 rad/s; radial velocity (r_dot) = 17.7000 ft/s, determine the transverse velocity (v_theta) in ft/s.

Start with the thinking

• The governing relation printed in this handbook section is Radial and transverse velocity components.
• Everything except v_theta is given, so isolate v_theta symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Radial and transverse components for planar motion describe a particle's velocity in polar coordinates.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

2. Step 2 — Rearrange symbolically for v_theta:

   $$v_{theta} = r \dot{\theta}$$

3. Step 3 — List the givens: radial distance (r) = 5.7000 ft, angular rate (theta_dot) = 2.4000 rad/s, radial velocity (r_dot) = 17.7000 ft/s.

4. Step 4 — Substitute the given values:

   $$v_{theta} = 5.7000 \dot{\theta}$$

5. Step 5 — Evaluate:

   $$v_{theta} = 13.6800\ \text{ft/s}$$

6. Step 6 — Check: returning v_theta = 13.6800 ft/s to

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 27.3600 — kept a factor of two that cancels in the correct rearrangement.
• 6.8400 — dropped that same factor in the other direction.
• 15.0480 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Radial and Transverse Components for Planar Motion

A particle sliding in a rotating slot is analyzed with radial and transverse components. Given angular rate (theta_dot) = 3.2000 rad/s; transverse velocity (v_theta) = 47.0000 ft/s; radial velocity (r_dot) = 11.9000 ft/s, determine the radial distance (r) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Radial and transverse velocity components.
• Everything except r is given, so isolate r symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Radial and transverse components for planar motion describe a particle's velocity in polar coordinates.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

2. Step 2 — Rearrange symbolically for r:

   $$r = \dfrac{v_\theta}{\dot{\theta}}$$

3. Step 3 — List the givens: angular rate (theta_dot) = 3.2000 rad/s, transverse velocity (v_theta) = 47.0000 ft/s, radial velocity (r_dot) = 11.9000 ft/s.

4. Step 4 — Substitute the given values:

   $$r = \dfrac{v_\theta}{\dot{\theta}}$$

5. Step 5 — Evaluate:

   $$r = 14.6875\ \text{ft}$$

6. Step 6 — Check: returning r = 14.6875 ft to

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 29.3750 — kept a factor of two that cancels in the correct rearrangement.
• 7.3438 — dropped that same factor in the other direction.
• 16.1563 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Radial and Transverse Components for Planar Motion

A satellite's planar motion is decomposed into radial and transverse components. Given radial distance (r) = 8.3000 ft; transverse velocity (v_theta) = 35.3000 ft/s; radial velocity (r_dot) = 16.7000 ft/s, determine the angular rate (theta_dot) in rad/s.

Start with the thinking

• The governing relation printed in this handbook section is Radial and transverse velocity components.
• Everything except theta_dot is given, so isolate theta_dot symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Radial and transverse components for planar motion describe a particle's velocity in polar coordinates.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

2. Step 2 — Rearrange symbolically for theta_dot:

   $$\theta_{dot} = \dfrac{v_\theta}{r}$$

3. Step 3 — List the givens: radial distance (r) = 8.3000 ft, transverse velocity (v_theta) = 35.3000 ft/s, radial velocity (r_dot) = 16.7000 ft/s.

4. Step 4 — Substitute the given values:

   $$\theta_{dot} = \dfrac{v_\theta}{8.3000}$$

5. Step 5 — Evaluate:

   $$\theta_{dot} = 4.2530\ \text{rad/s}$$

6. Step 6 — Check: returning theta_dot = 4.2530 rad/s to

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 8.5060 — kept a factor of two that cancels in the correct rearrangement.
• 2.1265 — dropped that same factor in the other direction.
• 4.6783 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Radial and Transverse Components for Planar Motion

A radar tracks an aircraft using radial and transverse velocity components. Given radial distance (r) = 9.7000 ft; angular rate (theta_dot) = 4.0000 rad/s; radial velocity (r_dot) = 6.8000 ft/s, determine the transverse velocity (v_theta) in ft/s.

Start with the thinking

• The governing relation printed in this handbook section is Radial and transverse velocity components.
• Everything except v_theta is given, so isolate v_theta symbolically first — never rearrange after the numbers are in.
• Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
• Radial and transverse components for planar motion describe a particle's velocity in polar coordinates.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

2. Step 2 — Rearrange symbolically for v_theta:

   $$v_{theta} = r \dot{\theta}$$

3. Step 3 — List the givens: radial distance (r) = 9.7000 ft, angular rate (theta_dot) = 4.0000 rad/s, radial velocity (r_dot) = 6.8000 ft/s.

4. Step 4 — Substitute the given values:

   $$v_{theta} = 9.7000 \dot{\theta}$$

5. Step 5 — Evaluate:

   $$v_{theta} = 38.8000\ \text{ft/s}$$

6. Step 6 — Check: returning v_theta = 38.8000 ft/s to

   $$v_r = \dot{r}, \quad v_\theta = r \dot{\theta}$$

   reproduces the given quantities, and both sides carry the same units.

Why the other options are there

• 77.6000 — kept a factor of two that cancels in the correct rearrangement.
• 19.4000 — dropped that same factor in the other direction.
• 42.6800 — rounded an intermediate value before the final step.

Reference: FE Handbook — Dynamics: Radial and Transverse Components for Planar Motion