A Car Travels Up A Hill At A Constant Speed Of 42 Km

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This assignment involves multiple physics and kinematics problems related to particle motion and average speeds. The tasks include calculating average speed for a round trip, determining times and positions when a particle stops or passes through the origin, analyzing position, velocity, and acceleration graphs, and solving equations describing motion along an x-axis. The primary focus is on understanding concepts of constant velocity, displacement, velocity, acceleration, maximums, and average values, supported by mathematical computations, derivatives, and kinematic equations.

Paper For Above instruction

The study of motion, especially particle kinematics, forms a foundational aspect of classical mechanics. Through analyzing various problems involving constant and variable velocity, maximum displacement, instantaneous stop points, and average velocity, we deepen our understanding of how objects move in a one-dimensional context. This paper systematically explores these themes through specific problems, with comprehensive calculations and interpretations based on principles of physics and mathematics.

Problem 1: Average speed of a round trip

A car travels up a hill at a constant speed of 42 km/h and then returns down the hill at 61 km/h. To find the average speed for the entire round trip, we employ the harmonic mean formula, which is appropriate when the distances are equal but the speeds differ. The formula for average speed v̅ when covering the same distance d at different speeds v₁ and v₂ is:

v̅ = 2 v₁ v₂ / (v₁ + v₂).

Substituting the given values:

v̅ = 2 42 61 / (42 + 61) = 5124 / 103 ≈ 49.74 km/h.

Considering a tolerance of +/-2%, the acceptable range of the average speed is approximately 48.75 km/h to 50.73 km/h. The calculated average, ~49.74 km/h, fits comfortably within this tolerance, indicating consistency and accuracy in the calculation.

Problem 2: Particle position, stop, and passing through origin

The particle's position function is given by x(t) = 5.00 - 7.00 t², with x in meters and t in seconds.

• (a) When does the particle stop? The particle stops when velocity is zero. Since velocity v(t) = dx/dt = -14 t, setting v(t) = 0 yields -14 t = 0 ⇒ t = 0. At t=0, the particle's initial position is x(0) = 5.00 meters.
• (b) Where does it stop? The position at t=0 is 5.00 meters.
• (c) When does the particle pass through the origin at negative time? Equate x(t) = 0:

  0 = 5.00 - 7.00 t² ⇒ 7 t² = 5.00 ⇒ t² = 5/7 ≈ 0.714 ⇒ t ≈ ±√0.714 ≈ ±0.845 seconds. The negative time is approximately -0.845 seconds.

• (d) When does the particle pass through the origin at positive time? At t ≈ +0.845 seconds.

Problem 3: Position, velocity, acceleration, and maximums

The position function is x(t) = 16.0 t² - 5.00 t³.

• (a) Position at t=7.00 s:

  x(7) = 16 49 - 5 343 = 784 - 1715 = -931 meters.

• (b) Velocity: Derivative of position:

  v(t) = dx/dt = 32.0 t - 15.0 t². Plug in t=7.00 s:

  v(7) = 32 7 - 15 49 = 224 - 735 = -511 m/s.

• (c) Acceleration: Derivative of velocity:

  a(t) = dv/dt = 32.0 - 30.0 t. At t=7.00 s:

  a(7) = 32 - 30 * 7 = 32 - 210 = -178 m/s².

• (d) Maximum positive coordinate:Find critical points by setting v(t) = 0:

  32 t - 15 t² = 0 ⇒ t (32 - 15 t) = 0 ⇒ t=0 or t = 32/15 ≈ 2.13 s. At t≈2.13 s, the position is:

  x(2.13) = 16(2.13)^2 - 5(2.13)^3 ≈ 164.54 - 59.66 ≈ 72.64 - 48.30 ≈ 24.34 meters. The maximum positive coordinate is approximately 24.34 meters reached at about 2.13 seconds.

• (e) Time when maximum coordinate is reached: Approximately 2.13 seconds.
• (f) Maximum positive velocity: Achieved at the critical point of velocity derivative, which occurs at t=0; maximum magnitude is at t=0: v(0)=0. The maximum positive velocity occurs when v(t) is maximized; since velocity is linear in t, the maximum occurs at t=0: 0 m/s. But analyzing v(t), its maximum positive occurs at t=0, confirming that the peak positive velocity is 0 at t=0.
• (g) Time at which maximum velocity is reached: At t=0 with velocity 0, but if considering the maximum absolute velocity, it occurs at t=7: v(7)=-511 m/s, which is negative. So, the maximum positive value is at t=0: 0 m/s.
• (h) Acceleration when the particle is momentarily stationary (other than at t=0): At t=2.13 s (when velocity=0), acceleration is:

  a(2.13) = 32 - 30*2.13 ≈ 32 - 63.9 ≈ -31.9 m/s².

• (i) The average velocity between t=0 and t=7.00 s:

  Average velocity = (x(7) - x(0)) / (7 - 0) = (-931 - 0) / 7 ≈ -133 m/s.

Problem 4: Position, displacement, velocity, and acceleration over a time interval

The position function is x(t) = c t⁴ - b t⁵ with c=2.5 m/s⁴ and b=1.4 m/s⁵.

• (a) Displacement from t=0 to t=2.1 s:

  x(2.1) = 2.5(2.1)^4 - 1.4(2.1)^5. Calculating:

  2.1^4 ≈ 19.4481, 2.1^5 ≈ 40.8521.

  Hence, x(2.1) ≈ 2.519.4481 - 1.440.8521 ≈ 48.62 - 57.11 ≈ -8.49 meters.

  • (b) Velocity at specific times:
  • At t=1.0 s: v = d/dt x(t) = 4c t³ - 5b t⁴.

    v(1) = 42.51 - 51.41 ≈ 10 - 7 = 3 m/s.

  • At t=2.0 s: v = 42.58 - 51.416 = 80 - 112 = -32 m/s.
  • At t=3.0 s: v = 42.527 - 51.481 ≈ 270 - 567 = -297 m/s.
  • At t=4.0 s: v = 42.564 - 51.4256 ≈ 640 - 1792 = -1152 m/s.
• (c) Acceleration at specific times:
• Acceleration is derivative of velocity: a(t)=d²x/dt² = 12 c t² - 20 b t³,So at t=1.0 s: a= 122.51 - 201.41 = 30 - 28 = 2 m/s².
• At t=2.0 s: a= 122.54 - 201.48 = 120 - 224 = -104 m/s².
• At t=3.0 s: a= 122.59 - 201.427 = 270 - 756 = -486 m/s².
• At t=4.0 s: a= 122.516 - 201.464 = 480 - 1792 = -1312 m/s².

These calculations demonstrate how displacement, velocity, and acceleration evolve over time for particles moving under polynomial position functions, illustrating critical points, maximum displacements, and typical motion characteristics.

Conclusion

Understanding particle motion involves analyzing position functions, calculating velocities and accelerations through derivatives, and interpreting maximum point behaviors. The problems discussed show that harmonic means effectively evaluate average speeds in round trips, while derivatives identify key motion characteristics such as stops and maxima. The use of calculus tools allows precise determination of particle behavior over time, thereby underpinning physical intuition in kinematic analysis. These methodologies form a core component of classical mechanics education and application.

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