A truck travelling at 60 mph decelerates at 11.2 ft/s². What distance does it need to stop?

Start with the thinking

• Convert mph to ft/s before anything else: 60 mph = 88 ft/s.
• Use the velocity-position relation to avoid solving for time.

Step-by-step solution

1. Convert

   $$v_{0} = 60(1.467) = 88.0 ft/s$$

2. Kinematics

   $$v^{2} = v_{0}^{2} + 2a d$$

3. Set v = 0

   $$0 = 88.0^{2} - 2(11.2)d$$

4. Rearrange

   $$d = 7,744/22.4$$

5. Result

   $$d = 346 ft$$

Why the other options are there

• 160 ft (mph used directly)
• 692 ft (factor of 2 dropped)

Reference: FE Reference Handbook — Dynamics — Rectilinear motion

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 95.0000 ft/s; acceleration (a) = 24.5000 ft/s²; distance (s) = 276.0 ft, determine the final velocity (v) in ft/s.

Start with the thinking

• The governing relation printed in this handbook section is Constant-acceleration kinematics.
• Everything except v is given, so isolate v 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.
• Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v^2 = v_0^2 + 2 a s$$

2. Step 2 — Rearrange the relation so that v stands alone on the left-hand side.

3. Step 3 — List the givens: initial velocity (v0) = 95.0000 ft/s, acceleration (a) = 24.5000 ft/s², distance (s) = 276.0 ft.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$v = 150.2\ \text{ft/s}$$

6. Step 6 — Check: returning v = 150.2 ft/s to

   $$v^2 = v_0^2 + 2 a s$$

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

Why the other options are there

• 300.3 — kept a factor of two that cancels in the correct rearrangement.
• 75.0816 — dropped that same factor in the other direction.
• 165.2 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Particle Kinematics

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 57.0000 ft/s; distance (s) = 460.0 ft; final velocity (v) = 83.7000 ft/s, determine the acceleration (a) in ft/s².

Start with the thinking

• The governing relation printed in this handbook section is Constant-acceleration kinematics.
• Everything except a is given, so isolate a 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.
• Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v^2 = v_0^2 + 2 a s$$

2. Step 2 — Rearrange the relation so that a stands alone on the left-hand side.

3. Step 3 — List the givens: initial velocity (v0) = 57.0000 ft/s, distance (s) = 460.0 ft, final velocity (v) = 83.7000 ft/s.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$a = 4.0834\ \text{ft/s²}$$

6. Step 6 — Check: returning a = 4.0834 ft/s² to

   $$v^2 = v_0^2 + 2 a s$$

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

Why the other options are there

• 8.1667 — kept a factor of two that cancels in the correct rearrangement.
• 2.0417 — dropped that same factor in the other direction.
• 4.4917 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Particle Kinematics

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 76.0000 ft/s; acceleration (a) = 13.5000 ft/s²; final velocity (v) = 48.1000 ft/s, determine the distance (s) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Constant-acceleration kinematics.
• Everything except s is given, so isolate s 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.
• Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v^2 = v_0^2 + 2 a s$$

2. Step 2 — Rearrange the relation so that s stands alone on the left-hand side.

3. Step 3 — List the givens: initial velocity (v0) = 76.0000 ft/s, acceleration (a) = 13.5000 ft/s², final velocity (v) = 48.1000 ft/s.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$s = -128.2\ \text{ft}$$

6. Step 6 — Check: returning s = -128.2 ft to

   $$v^2 = v_0^2 + 2 a s$$

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

Why the other options are there

• -256.5 — kept a factor of two that cancels in the correct rearrangement.
• -64.1183 — dropped that same factor in the other direction.
• -141.1 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Particle Kinematics

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 53.0000 ft/s; acceleration (a) = -6.0000 ft/s²; distance (s) = 166.0 ft, determine the final velocity (v) in ft/s.

Start with the thinking

• The governing relation printed in this handbook section is Constant-acceleration kinematics.
• Everything except v is given, so isolate v 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.
• Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v^2 = v_0^2 + 2 a s$$

2. Step 2 — Rearrange the relation so that v stands alone on the left-hand side.

3. Step 3 — List the givens: initial velocity (v0) = 53.0000 ft/s, acceleration (a) = -6.0000 ft/s², distance (s) = 166.0 ft.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$v = 28.5832\ \text{ft/s}$$

6. Step 6 — Check: returning v = 28.5832 ft/s to

   $$v^2 = v_0^2 + 2 a s$$

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

Why the other options are there

• 57.1664 — kept a factor of two that cancels in the correct rearrangement.
• 14.2916 — dropped that same factor in the other direction.
• 31.4415 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Particle Kinematics

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 39.0000 ft/s; distance (s) = 373.0 ft; final velocity (v) = 72.0000 ft/s, determine the acceleration (a) in ft/s².

Start with the thinking

• The governing relation printed in this handbook section is Constant-acceleration kinematics.
• Everything except a is given, so isolate a 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.
• Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v^2 = v_0^2 + 2 a s$$

2. Step 2 — Rearrange the relation so that a stands alone on the left-hand side.

3. Step 3 — List the givens: initial velocity (v0) = 39.0000 ft/s, distance (s) = 373.0 ft, final velocity (v) = 72.0000 ft/s.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$a = 4.9102\ \text{ft/s²}$$

6. Step 6 — Check: returning a = 4.9102 ft/s² to

   $$v^2 = v_0^2 + 2 a s$$

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

Why the other options are there

• 9.8204 — kept a factor of two that cancels in the correct rearrangement.
• 2.4551 — dropped that same factor in the other direction.
• 5.4012 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Particle Kinematics

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 26.0000 ft/s; acceleration (a) = -18.0000 ft/s²; final velocity (v) = 55.1000 ft/s, determine the distance (s) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Constant-acceleration kinematics.
• Everything except s is given, so isolate s 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.
• Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v^2 = v_0^2 + 2 a s$$

2. Step 2 — Rearrange the relation so that s stands alone on the left-hand side.

3. Step 3 — List the givens: initial velocity (v0) = 26.0000 ft/s, acceleration (a) = -18.0000 ft/s², final velocity (v) = 55.1000 ft/s.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$s = -65.5558\ \text{ft}$$

6. Step 6 — Check: returning s = -65.5558 ft to

   $$v^2 = v_0^2 + 2 a s$$

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

Why the other options are there

• -131.1 — kept a factor of two that cancels in the correct rearrangement.
• -32.7779 — dropped that same factor in the other direction.
• -72.1114 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Particle Kinematics

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 78.0000 ft/s; acceleration (a) = 17.0000 ft/s²; distance (s) = 359.0 ft, determine the final velocity (v) in ft/s.

Start with the thinking

• The governing relation printed in this handbook section is Constant-acceleration kinematics.
• Everything except v is given, so isolate v 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.
• Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v^2 = v_0^2 + 2 a s$$

2. Step 2 — Rearrange the relation so that v stands alone on the left-hand side.

3. Step 3 — List the givens: initial velocity (v0) = 78.0000 ft/s, acceleration (a) = 17.0000 ft/s², distance (s) = 359.0 ft.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$v = 135.2\ \text{ft/s}$$

6. Step 6 — Check: returning v = 135.2 ft/s to

   $$v^2 = v_0^2 + 2 a s$$

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

Why the other options are there

• 270.5 — kept a factor of two that cancels in the correct rearrangement.
• 67.6203 — dropped that same factor in the other direction.
• 148.8 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Particle Kinematics

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 49.0000 ft/s; distance (s) = 383.0 ft; final velocity (v) = 32.8000 ft/s, determine the acceleration (a) in ft/s².

Start with the thinking

• The governing relation printed in this handbook section is Constant-acceleration kinematics.
• Everything except a is given, so isolate a 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.
• Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v^2 = v_0^2 + 2 a s$$

2. Step 2 — Rearrange the relation so that a stands alone on the left-hand side.

3. Step 3 — List the givens: initial velocity (v0) = 49.0000 ft/s, distance (s) = 383.0 ft, final velocity (v) = 32.8000 ft/s.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$a = -1.7300\ \text{ft/s²}$$

6. Step 6 — Check: returning a = -1.7300 ft/s² to

   $$v^2 = v_0^2 + 2 a s$$

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

Why the other options are there

• -3.4599 — kept a factor of two that cancels in the correct rearrangement.
• -0.8650 — dropped that same factor in the other direction.
• -1.9030 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Particle Kinematics

A dynamics problem uses Constant-acceleration kinematics. Given initial velocity (v0) = 100.0 ft/s; acceleration (a) = 19.5000 ft/s²; final velocity (v) = 84.6000 ft/s, determine the distance (s) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Constant-acceleration kinematics.
• Everything except s is given, so isolate s 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.
• Dynamics items reward recognising the unknown before touching a calculator.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$v^2 = v_0^2 + 2 a s$$

2. Step 2 — Rearrange the relation so that s stands alone on the left-hand side.

3. Step 3 — List the givens: initial velocity (v0) = 100.0 ft/s, acceleration (a) = 19.5000 ft/s², final velocity (v) = 84.6000 ft/s.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$s = -72.8933\ \text{ft}$$

6. Step 6 — Check: returning s = -72.8933 ft to

   $$v^2 = v_0^2 + 2 a s$$

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

Why the other options are there

• -145.8 — kept a factor of two that cancels in the correct rearrangement.
• -36.4467 — dropped that same factor in the other direction.
• -80.1827 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Dynamics → Particle Kinematics