During A Hard Sneeze, Your Eyes Might Shut For A Few Seconds

During A Hard Sneeze Your Eyes Might Shut For056 S If You Are Dri

Analyze a series of physics problems related to motion, velocity, acceleration, and position functions of particles and vehicles. The focus includes calculating distances during short time intervals, average speeds over non-uniform trips, specific moments when particles stop or pass through particular points, and the maximum or minimum values of position, velocity, and acceleration. Consider different given functions for position and velocity, and apply calculus principles such as differentiation, as well as kinematic equations, to solve for various parameters. Incorporate tolerance levels where specified, and present comprehensive solutions to each subproblem, including calculating times, positions, velocities, accelerations, and displacements, while interpreting the physical context of each problem.

Paper For Above instruction

Understanding motion in physics involves analyzing how objects move along a path over time, which can be quantified using position functions, velocity, and acceleration. The problems presented examine various scenarios illustrating fundamental concepts in kinematics, such as displacement during a brief interval, average speed over trips with varying velocities, and specific instances when particles or objects pass through certain points or change their motion characteristics. Here, each problem is addressed methodically, applying relevant equations from classical mechanics, calculus, and algebra to derive precise numerical responses, considering the given constraints and tolerances.

1. Distance Traveled by a Car During a Sneeze

The problem describes a scenario where a person sneezes, causing their eyes to shut for 0.56 seconds, during which a car moving at 71 km/h travels a certain distance. To find this distance, the first step is to convert the velocity into meters per second: 71 km/h equals approximately 19.72 m/s (since 1 km/h ≈ 0.27778 m/s). The distance traveled during the sneeze duration can then be calculated using the relation: distance = velocity × time.

Applying the formula: distance = 19.72 m/s × 0.56 s ≈ 11.04 meters. Considering a tolerance of ±2%, the allowable variation in this distance is approximately ±0.22 meters, resulting in a range of roughly 10.82 to 11.26 meters.

This calculation demonstrates how shortly during a sneeze, a vehicle covers a significant distance, illustrating the importance of quick reflexes and awareness when driving under sudden causal events.

2. Average Speed for Round Trip over Hilly Terrain

In this problem, a car ascends a hill at 38 km/h and descends back at 59 km/h. To compute the average speed for the entire round trip, one must consider the different speeds and distances assumed to be equal (or specify otherwise). The average speed (v_avg) for two-part journeys with different speeds but equal distances (d) is given by:

v_avg = 2d / (t_up + t_down), where t_up = d / v_up and t_down = d / v_down.

Expressed explicitly, v_avg = 2d / (d / v_up + d / v_down) = 2 / (1 / v_up + 1 / v_down).

Converting speeds to meters per second: 38 km/h ≈ 10.56 m/s, and 59 km/h ≈ 16.39 m/s. Using these, the average speed becomes:

v_avg = 2 / (1 / 10.56 + 1 / 16.39) ≈ 2 / (0.0947 + 0.0610) ≈ 2 / 0.1557 ≈ 12.86 m/s.

Converting back to km/h: 12.86 m/s × 3.6 ≈ 46.3 km/h. Considering a ±2% tolerance, the possible average speed ranges from approximately 45.2 km/h to 47.4 km/h.

This calculation highlights how differing speeds over equal segments influence the overall trip average, crucial in trip planning and fuel consumption estimations.

3. Position Function and Particle Dynamics

The position function x(t) = 6.00 - 7.00 t² describes a particle moving along a line, with x in meters and t in seconds. To find when the particle stops, set velocity v(t) = dx/dt = -14.00 t, equal to zero, which occurs at t = 0 seconds. Additionally, the particle's position at that time is x(0) = 6.00 meters.

For negative and positive times when the particle passes through the origin (x=0), solve: 0 = 6.00 - 7.00 t²; so t² = 6.00/7.00 ≈ 0.857; t ≈ ±√0.857 ≈ ±0.926 s. The negative time is approximately -0.926 seconds, and the positive time is approximately +0.926 seconds.

These calculations help determine when the particle is at rest or passes specific points, which involves understanding motion symmetry and quadratic position functions.

4. Particle Motion with Given Position Function

The position function x(t) = 13.0 t² - 3.00 t³ allows analysis of maximums and points of zero velocity. The velocity v(t) = dx/dt = 26.0 t - 9.00 t²; setting v(t) = 0 yields t = 0 or t ≈ 26.0 / 9.00 ≈ 2.89 s for maximum velocity points.

At t=7.00 s: position x(7) = 13.0×49 - 3.00×343 = 637 - 1029 = -392 meters; velocity v(7) = 26.0×7 - 9.00×49 = 182 - 441 = -259 m/s; acceleration a(t) = dv/dt = 26.0 - 18.00 t, thus a(7) = 26 - 126 = -100 m/s².

The maximum positive position occurs at critical points where dv/dt changes sign; maximum position is at t ≈ 2.89 s, with x ≈ x(2.89) ≈ 13×8.34 - 3×24.2 ≈ 108.4 meters. Maximum velocity is at t=2.89 s: v ≈ 26×2.89 - 9×8.34 ≈ 75.14 - 75.06 ≈ 0.08 m/s, effectively near zero, indicating a turning point. At the instant when the particle is at rest (other than t=0), it occurs at t ≈ 2.89 s.

The average velocity between t=0 and t=7s is (x(7) - x(0)) / 7 ≈ (-392 - 0) / 7 ≈ -56 m/s, indicating motion towards the negative x-direction over that interval.

5. Motion with Power-Law Position Function

The position function x(t) = c t⁵ - b t⁶ with c=2.4 m/s⁵ and b=1.6 m/s⁶ describes a particle's displacement. The displacement from t=0 to t=1.9 s is:

Δx = x(1.9) - x(0) = 2.4×(1.9)^5 - 1.6×(1.9)^6 ≈ 2.4×24.76 - 1.6×47.04 ≈ 59.45 - 75.26 ≈ -15.81 meters.

Velocity at specific times can be found using v(t) = dx/dt = 5c t⁴ - 6b t⁵. At t=1.0 s: v(1) = 5×2.4×1 - 6×1.6×1 ≈ 12 - 9.6 = 2.4 m/s. At t=2.0 s: v(2) = 5×2.4×16 - 6×1.6×32 ≈ 192 - 307.2 ≈ -115.2 m/s. At t=3.0 s: v(3) = 5×2.4×81 - 6×1.6×243 ≈ 972 - 2332. ≈ -1360 m/s. At t=4.0 s: v(4) = 5×2.4×256 - 6×1.6×1024 ≈ 3072 - 9830.4 ≈ -6758.4 m/s. The acceleration a(t) = d²x/dt² = 20c t³ - 30b t⁴.

At t=1.0 s: a(1) ≈ 20×2.4×1 - 30×1.6×1 ≈ 48 - 48 = 0 m/s². At t=2.0 s: a(2) ≈ 20×2.4×8 - 30×1.6×16 ≈ 384 - 768 ≈ -384 m/s². At t=3.0 s: a(3) ≈ 20×2.4×27 - 30×1.6×81 ≈ 1296 - 3888 ≈ -2592 m/s². At t=4.0 s: a(4) ≈ 20×2.4×64 - 30×1.6×256 ≈ 3072 - 12288 ≈ -9216 m/s². These calculations show how both velocity and acceleration evolve over time, indicating dynamic motion with initial acceleration, followed by rapid deceleration.

References

• Serway, R. A., & Jewett, J. W. (2014). Physics for Scientists and Engineers. Brooks Cole.
• Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. Wiley.
• Young, H. D., Freedman, R. A. (2019). Sears & Zemansky's University Physics. Pearson.
• Tipler, P. A., & Mosca, G. (2008). Physics for Scientists and Engineers. W.H. Freeman.
• Knight, R. D. (2012). Physics for Scientists and Engineers. Pearson.
• Hibbeler, R. C. (2011). Engineering Mechanics: Dynamics. Pearson.
• Koenigsberger, E. (2011). Classical Mechanics. InTech Open.
• Franklin, J. (2017). Classical Dynamics. Springer.
• Ginsberg, N. S. (2001). Mechanical and Electrical Oscillations. CRC Press.
• Chabay, R., & Sherwood, B. (2011). Matter & Interactions. Wiley.