A vehicle moving in a straight line has an initial speed of 26 m/s and a constant acceleration of 3.0 m/s². Using the equations of motion in Cartesian coordinates, find the velocity and position after 10 s, and the speed after travelling 200 m.

Start with the thinking

• With constant acceleration the Cartesian equations of motion apply directly — no integration is needed.
• The velocity-position relation avoids solving for time.

Step-by-step solution

1. Formula

   $$v = v_{0} + a t$$

2. Substituting

   $$v = 26 + 3.0(10) = 56.00 m/s$$

3. Formula

   $$s = v_{0} t + ½ a t^{2}$$

4. Substituting

   $$s = 26(10) + ½(3.0)(10)^{2} = 410.0 m$$

5. Formula

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

6. Substituting

   $$v^{2} = 26^{2} + 2(3.0)(200) = 1,876$$

7. Evaluate

   $$v = 43.31 m/s$$

Why the other options are there

• s = 260.0 m (acceleration term dropped)
• v = 30.00 m/s (initial speed dropped)

Reference: FE Reference Handbook — Dynamics → Cartesian Coordinates

A vehicle moving in a straight line has an initial speed of 27 m/s and a constant acceleration of 1.5 m/s². Using the equations of motion in Cartesian coordinates, find the velocity and position after 11 s, and the speed after travelling 200 m.

Start with the thinking

• With constant acceleration the Cartesian equations of motion apply directly — no integration is needed.
• The velocity-position relation avoids solving for time.

Step-by-step solution

1. Formula

   $$v = v_{0} + a t$$

2. Substituting

   $$v = 27 + 1.5(11) = 43.50 m/s$$

3. Formula

   $$s = v_{0} t + ½ a t^{2}$$

4. Substituting

   $$s = 27(11) + ½(1.5)(11)^{2} = 387.8 m$$

5. Formula

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

6. Substituting

   $$v^{2} = 27^{2} + 2(1.5)(200) = 1,329$$

7. Evaluate

   $$v = 36.46 m/s$$

Why the other options are there

• s = 297.0 m (acceleration term dropped)
• v = 16.50 m/s (initial speed dropped)

Reference: FE Reference Handbook — Dynamics → Cartesian Coordinates

A vehicle moving in a straight line has an initial speed of 28 m/s and a constant acceleration of 2.5 m/s². Using the equations of motion in Cartesian coordinates, find the velocity and position after 10 s, and the speed after travelling 200 m.

Start with the thinking

• With constant acceleration the Cartesian equations of motion apply directly — no integration is needed.
• The velocity-position relation avoids solving for time.

Step-by-step solution

1. Formula

   $$v = v_{0} + a t$$

2. Substituting

   $$v = 28 + 2.5(10) = 53.00 m/s$$

3. Formula

   $$s = v_{0} t + ½ a t^{2}$$

4. Substituting

   $$s = 28(10) + ½(2.5)(10)^{2} = 405.0 m$$

5. Formula

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

6. Substituting

   $$v^{2} = 28^{2} + 2(2.5)(200) = 1,784$$

7. Evaluate

   $$v = 42.24 m/s$$

Why the other options are there

• s = 280.0 m (acceleration term dropped)
• v = 25.00 m/s (initial speed dropped)

Reference: FE Reference Handbook — Dynamics → Cartesian Coordinates

A vehicle moving in a straight line has an initial speed of 25 m/s and a constant acceleration of 3.0 m/s². Using the equations of motion in Cartesian coordinates, find the velocity and position after 9 s, and the speed after travelling 200 m.

Start with the thinking

• With constant acceleration the Cartesian equations of motion apply directly — no integration is needed.
• The velocity-position relation avoids solving for time.

Step-by-step solution

1. Formula

   $$v = v_{0} + a t$$

2. Substituting

   $$v = 25 + 3.0(9) = 52.00 m/s$$

3. Formula

   $$s = v_{0} t + ½ a t^{2}$$

4. Substituting

   $$s = 25(9) + ½(3.0)(9)^{2} = 346.5 m$$

5. Formula

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

6. Substituting

   $$v^{2} = 25^{2} + 2(3.0)(200) = 1,825$$

7. Evaluate

   $$v = 42.72 m/s$$

Why the other options are there

• s = 225.0 m (acceleration term dropped)
• v = 27.00 m/s (initial speed dropped)

Reference: FE Reference Handbook — Dynamics → Cartesian Coordinates

A vehicle moving in a straight line has an initial speed of 5 m/s and a constant acceleration of 3.5 m/s². Using the equations of motion in Cartesian coordinates, find the velocity and position after 9 s, and the speed after travelling 200 m.

Start with the thinking

• With constant acceleration the Cartesian equations of motion apply directly — no integration is needed.
• The velocity-position relation avoids solving for time.

Step-by-step solution

1. Formula

   $$v = v_{0} + a t$$

2. Substituting

   $$v = 5 + 3.5(9) = 36.50 m/s$$

3. Formula

   $$s = v_{0} t + ½ a t^{2}$$

4. Substituting

   $$s = 5(9) + ½(3.5)(9)^{2} = 186.8 m$$

5. Formula

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

6. Substituting

   $$v^{2} = 5^{2} + 2(3.5)(200) = 1,425$$

7. Evaluate

   $$v = 37.75 m/s$$

Why the other options are there

• s = 45.0 m (acceleration term dropped)
• v = 31.50 m/s (initial speed dropped)

Reference: FE Reference Handbook — Dynamics → Cartesian Coordinates

A vehicle moving in a straight line has an initial speed of 16 m/s and a constant acceleration of 3.5 m/s². Using the equations of motion in Cartesian coordinates, find the velocity and position after 8 s, and the speed after travelling 200 m.

Start with the thinking

• With constant acceleration the Cartesian equations of motion apply directly — no integration is needed.
• The velocity-position relation avoids solving for time.

Step-by-step solution

1. Formula

   $$v = v_{0} + a t$$

2. Substituting

   $$v = 16 + 3.5(8) = 44.00 m/s$$

3. Formula

   $$s = v_{0} t + ½ a t^{2}$$

4. Substituting

   $$s = 16(8) + ½(3.5)(8)^{2} = 240.0 m$$

5. Formula

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

6. Substituting

   $$v^{2} = 16^{2} + 2(3.5)(200) = 1,656$$

7. Evaluate

   $$v = 40.69 m/s$$

Why the other options are there

• s = 128.0 m (acceleration term dropped)
• v = 28.00 m/s (initial speed dropped)

Reference: FE Reference Handbook — Dynamics → Cartesian Coordinates

A vehicle moving in a straight line has an initial speed of 23 m/s and a constant acceleration of 0.5 m/s². Using the equations of motion in Cartesian coordinates, find the velocity and position after 7 s, and the speed after travelling 200 m.

Start with the thinking

• With constant acceleration the Cartesian equations of motion apply directly — no integration is needed.
• The velocity-position relation avoids solving for time.

Step-by-step solution

1. Formula

   $$v = v_{0} + a t$$

2. Substituting

   $$v = 23 + 0.5(7) = 26.50 m/s$$

3. Formula

   $$s = v_{0} t + ½ a t^{2}$$

4. Substituting

   $$s = 23(7) + ½(0.5)(7)^{2} = 173.3 m$$

5. Formula

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

6. Substituting

   $$v^{2} = 23^{2} + 2(0.5)(200) = 729.0$$

7. Evaluate

   $$v = 27.00 m/s$$

Why the other options are there

• s = 161.0 m (acceleration term dropped)
• v = 3.50 m/s (initial speed dropped)

Reference: FE Reference Handbook — Dynamics → Cartesian Coordinates

A vehicle moving in a straight line has an initial speed of 17 m/s and a constant acceleration of 1.5 m/s². Using the equations of motion in Cartesian coordinates, find the velocity and position after 11 s, and the speed after travelling 200 m.

Start with the thinking

• With constant acceleration the Cartesian equations of motion apply directly — no integration is needed.
• The velocity-position relation avoids solving for time.

Step-by-step solution

1. Formula

   $$v = v_{0} + a t$$

2. Substituting

   $$v = 17 + 1.5(11) = 33.50 m/s$$

3. Formula

   $$s = v_{0} t + ½ a t^{2}$$

4. Substituting

   $$s = 17(11) + ½(1.5)(11)^{2} = 277.8 m$$

5. Formula

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

6. Substituting

   $$v^{2} = 17^{2} + 2(1.5)(200) = 889.0$$

7. Evaluate

   $$v = 29.82 m/s$$

Why the other options are there

• s = 187.0 m (acceleration term dropped)
• v = 16.50 m/s (initial speed dropped)

Reference: FE Reference Handbook — Dynamics → Cartesian Coordinates

A vehicle moving in a straight line has an initial speed of 12 m/s and a constant acceleration of 1.5 m/s². Using the equations of motion in Cartesian coordinates, find the velocity and position after 8 s, and the speed after travelling 200 m.

Start with the thinking

• With constant acceleration the Cartesian equations of motion apply directly — no integration is needed.
• The velocity-position relation avoids solving for time.

Step-by-step solution

1. Formula

   $$v = v_{0} + a t$$

2. Substituting

   $$v = 12 + 1.5(8) = 24.00 m/s$$

3. Formula

   $$s = v_{0} t + ½ a t^{2}$$

4. Substituting

   $$s = 12(8) + ½(1.5)(8)^{2} = 144.0 m$$

5. Formula

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

6. Substituting

   $$v^{2} = 12^{2} + 2(1.5)(200) = 744.0$$

7. Evaluate

   $$v = 27.28 m/s$$

Why the other options are there

• s = 96.0 m (acceleration term dropped)
• v = 12.00 m/s (initial speed dropped)

Reference: FE Reference Handbook — Dynamics → Cartesian Coordinates

A vehicle moving in a straight line has an initial speed of 26 m/s and a constant acceleration of 1.0 m/s². Using the equations of motion in Cartesian coordinates, find the velocity and position after 12 s, and the speed after travelling 200 m.

Start with the thinking

• With constant acceleration the Cartesian equations of motion apply directly — no integration is needed.
• The velocity-position relation avoids solving for time.

Step-by-step solution

1. Formula

   $$v = v_{0} + a t$$

2. Substituting

   $$v = 26 + 1.0(12) = 38.00 m/s$$

3. Formula

   $$s = v_{0} t + ½ a t^{2}$$

4. Substituting

   $$s = 26(12) + ½(1.0)(12)^{2} = 384.0 m$$

5. Formula

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

6. Substituting

   $$v^{2} = 26^{2} + 2(1.0)(200) = 1,076$$

7. Evaluate

   $$v = 32.80 m/s$$

Why the other options are there

• s = 312.0 m (acceleration term dropped)
• v = 12.00 m/s (initial speed dropped)

Reference: FE Reference Handbook — Dynamics → Cartesian Coordinates