Work Out The Following Problems Be Sure To Show Your Work

Work Out The Following Problems Be Sure To Show Your Work In Detail

Identify, analyze, and solve each of the given physics and mathematics problems step-by-step, providing clear explanations and detailed calculations for each. Use appropriate formulas, units, and logical reasoning to arrive at accurate solutions. Show all work comprehensively to demonstrate understanding and problem-solving skills.

Paper For Above instruction

Problem 1: Displacement of a car between Oklahoma City and Tulsa

The problem states that a car is traveling along a straight highway between Oklahoma City and Tulsa. At one point, it is 50 miles from Tulsa, and at a later point, it is 40 miles from Tulsa. The goal is to determine the car’s displacement.

Displacement is a vector quantity that points from the initial position to the final position, regardless of the path traveled. To solve this, we need to define a coordinate system and assign initial and final positions along that axis. Let's assume Tulsa as the reference point at position zero.

Initial position: 50 miles from Tulsa.

Final position: 40 miles from Tulsa.

Displacement, Δx = final position - initial position = 40 miles - 50 miles = -10 miles.

The negative sign indicates the car moved in the direction toward Tulsa (assuming positive is away from Tulsa). Therefore, the displacement of the car is 10 miles toward Tulsa.

Answer: The displacement is 10 miles toward Tulsa, represented as -10 miles if considering direction.

Problem 2: Position of an ant after crawling on a coordinate plane

An ant crawls on graph paper with three segments:

• 3 cm along an azimuth (course angle) of 30°
• 5 cm along azimuth 180°
• 2 cm along azimuth 270°

Find the ant’s position relative to its starting point, in an X-Y coordinate system.

Assuming the starting point is at (0,0), we convert each movement into Cartesian components using basic trigonometry:

• For each segment, dx = length × cos(angle), dy = length × sin(angle), where the angles are measured from the positive X-axis, clockwise being negative, or according to standard mathematical conventions (counterclockwise from the positive X-axis). Since azimuths are typically measured clockwise from North, we need to convert them to standard mathematical angles:

Azimuth of 30° from North corresponds to an angle of 90° - 30° = 60° from the positive X-axis.

Similarly, azimuth 180° (from North) corresponds to 90° - 180° = -90° (or 270°) from the positive X-axis.

Azimuth 270° corresponds to 90° - 270° = -180°, which is equivalent to 180°, but considering movement along the axes, it matches moving south.

Calculations:

• First segment (3 cm at 30° azimuth):

  x₁ = 3 × cos(60°) = 3 × 0.5 = 1.5 cm

  y₁ = 3 × sin(60°) = 3 × (√3/2) ≈ 3 × 0.866 ≈ 2.598 cm

• Second segment (5 cm at 180° azimuth):

  x₂ = 5 × cos(-90°) = 5 × 0 = 0 cm

  y₂ = 5 × sin(-90°) = 5 × (-1) = -5 cm

• Third segment (2 cm at 270° azimuth):

  x₃ = 2 × cos(-180°) = 2 × (-1) = -2 cm

  y₃ = 2 × sin(-180°) = 2 × 0 = 0 cm

Adding components:

X total: 1.5 + 0 - 2 = -0.5 cm

Y total: 2.598 - 5 + 0 = -2.402 cm

The position of the ant with respect to its starting point is approximately (-0.5 cm, -2.402 cm).

Answer: The ant's final position is about 0.5 cm west and 2.4 cm south of the starting point.

Problem 3: Average velocity of a motorcycle between two points

A motorcycle covers a distance of 100 meters between two points in 1.2 seconds. Determine its average velocity.

Average velocity (v̄) = total displacement / time.

Since the points are straight and 100 meters apart, the displacement magnitude is 100 meters, in the direction of travel.

Assuming straight-line motion:

v̄ = 100 m / 1.2 s ≈ 83.33 m/s

Therefore, the average velocity magnitude is approximately 83.33 meters per second in the direction from the first to the second point.

Answer: The average velocity is approximately 83.33 m/s.

Problem 4: Velocity and acceleration of a bullet in a pistol barrel

a. Average velocity while in the barrel

The bullet travels 13 cm (0.13 m) and leaves the barrel at 300 m/s.

Average velocity (v_avg) in the barrel = total distance / time in the barrel.

We can find the time using the final velocity and assuming constant acceleration:

b. Time taken in the barrel assuming constant acceleration

Using the equation v_f = v_i + a t, where v_i ≈ 0 (assuming negligible initial velocity within the barrel), and v_f=300 m/s:

From calculus, average velocity under constant acceleration is:

v_avg = (v_i + v_f) / 2 = (0 + 300) / 2 = 150 m/s

Time in barrel: t = distance / v_avg = 0.13 m / 150 m/s ≈ 0.000867 s

Note: Alternatively, using v_f = a t, and v_i = 0, then a = v_f / t, and average velocity = (v_i + v_f)/2.

Calculating acceleration:

a = (v_f - v_i) / t = 300 / 0.000867 ≈ 345,732 m/s²

Alternatively, directly use v_avg to find time as above.

Answer (a & b): The average velocity during the bullet's passage in the barrel is approximately 150 m/s, and it takes roughly 0.000867 seconds to travel the 13 cm length of the barrel.

c. Average acceleration

a = (v_f - v_i) / t = (300 - 0) / 0.000867 ≈ 345,732 m/s²

Problem 5: Distance traveled by a missile propelled from rest with acceleration

The missile is propelled at 20 m/s initially, then accelerates at 20 g (where g ≈ 9.81 m/s²) after launch.

Initial velocity, v₀ = 20 m/s.

Acceleration, a = 20 g = 20 × 9.81 m/s² ≈ 196.2 m/s².

Time to hit target: t = 2 s, with ignition at t = 0.2 s.

We need to compute the total distance covered before and after ignition separately and then sum for total distance.

• Distance before ignition (0 to 0.2 s):

  Use: s = v₀ t + 0.5 a t²

  Note: Before ignition, acceleration is zero, so:

  s₁ = v₀ × t = 20 m/s × 0.2 s = 4 m

• Distance during acceleration (after ignition, from t=0.2 s to t=2 s):

  Total time after ignition: 2 - 0.2 = 1.8 s

  Initial velocity at ignition: v₀ = 20 m/s

  Using: s = v₀ t + 0.5 a t²

  s₂ = 20 m/s × 1.8 s + 0.5 × 196.2 m/s² × (1.8 s)²

  Calculate:

  s₂ = 36 m + 0.5 × 196.2 × 3.24 ≈ 36 + 0.5 × 196.2 × 3.24

  s₂ ≈ 36 + 0.5 × 196.2 × 3.24 ≈ 36 + 100.8 × 3.24 ≈ 36 + 327.0 ≈ 363.0 m

Adding both distances:

Total distance: s_total = s₁ + s₂ ≈ 4 + 363 ≈ 367 meters.

Thus, the target was approximately 367 meters away from the launch point.

References

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