🔑:You're attempting to describe motion using only spatial and kinematic quantities, without explicitly referencing time. This approach is indeed possible, and it's related to the concept of "geometric" or "kinematic" descriptions of motion.In your example, you're proposing to describe the motion by stating that the body moves 10 meters with a speed of 10V, where V is a fundamental unit of speed. This approach is similar to how we might describe a motion in terms of its spatial trajectory and velocity vector.To make this work, you would need to define a system of units where speed is a fundamental quantity, rather than a derived one. This is possible, but it would require a different set of definitions and conventions than the standard ones used in physics.In the standard SI system, speed is defined as the rate of change of position with respect to time (v = Δx / Δt). By defining speed as a fundamental unit, you would essentially be inverting this relationship, using speed as a primitive concept to describe motion, rather than deriving it from position and time.There are a few ways to approach this:1. Geometric algebra: This mathematical framework allows you to describe geometric and kinematic quantities, including motion, using a unified language. In geometric algebra, you can represent motion as a combination of spatial and directional quantities, without explicitly referencing time.2. Differential geometry: This branch of mathematics studies the properties of curves and surfaces in space. You can use differential geometry to describe motion as a curve in space, parameterized by a scalar parameter (e.g., arc length) rather than time.3. Non-standard analysis: This mathematical framework allows you to work with infinitesimal and infinite quantities in a rigorous way. You can use non-standard analysis to define a system of units where speed is a fundamental quantity, and then describe motion in terms of these units.While it's possible to develop a framework for describing motion without explicit reference to time, it's essential to note that:* Time is a fundamental aspect of our experience and is closely tied to the concept of causality.* The standard definitions of physical quantities, including speed, are based on the notion of time as a separate, one-dimensional quantity.* Alternative frameworks, like those mentioned above, may require significant changes to our understanding of physical concepts and the mathematical structures used to describe them.In summary, while it's theoretically possible to describe motion using only spatial and kinematic quantities, without explicit reference to time, it would require a significant reworking of our mathematical and conceptual frameworks. The standard approach, which uses time as a fundamental quantity, remains the most widely accepted and practical way to describe motion in physics.

🔑:## Step 1: Recall the equation of motion for an object under uniform acceleration.The equation of motion that relates displacement (x), initial velocity (v_0), acceleration (a), and time (t) is given by (x = v_0t + frac{1}{2}at^2).## Step 2: Substitute the given values into the equation of motion.Given that (v_0 = 0), (a = 12 , text{m/s}^2), and (x = 100 , text{m}), we substitute these into the equation to get (100 = 0 cdot t + frac{1}{2} cdot 12 cdot t^2).## Step 3: Simplify and solve the equation for (t).The equation simplifies to (100 = 6t^2). To solve for (t), we divide both sides by 6, resulting in (t^2 = frac{100}{6}). Taking the square root of both sides gives (t = sqrt{frac{100}{6}}).## Step 4: Calculate the numerical value of (t).(t = sqrt{frac{100}{6}} = sqrt{frac{50}{3}} approx sqrt{16.67} approx 4.08 , text{s}).## Step 5: Use the equation (v = v_0 + at) to find the final velocity (v).Given (v_0 = 0) and (a = 12 , text{m/s}^2), and using the calculated value of (t approx 4.08 , text{s}), we find (v = 0 + 12 cdot 4.08).## Step 6: Calculate the numerical value of (v).(v approx 12 cdot 4.08 approx 48.96 , text{m/s}).The final answer is: boxed{4.08}

🔑:To find the reading on the scale, which represents the apparent weight of the passenger, we need to consider the forces acting on the passenger. The primary forces are the passenger's weight (due to gravity) and the normal force exerted by the scale (which is the reading on the scale). When the elevator accelerates downward, the normal force (and thus the apparent weight) is less than the passenger's true weight because the elevator's downward acceleration reduces the force exerted by the scale on the passenger.Given:- Mass of the passenger (m) = 72.2 kg- Acceleration of the elevator (a) = 3.20 m/s^2 downward- Acceleration due to gravity (g) = 9.80 m/s^2 downwardThe formula to find the apparent weight (W_app) when the elevator is accelerating is:[ W_{app} = m(g - a) ]for downward acceleration, or[ W_{app} = m(g + a) ]for upward acceleration. Since the elevator is going down, we use the first formula.Substituting the given values:[ W_{app} = 72.2 , text{kg} times (9.80 , text{m/s}^2 - 3.20 , text{m/s}^2) ][ W_{app} = 72.2 , text{kg} times 6.60 , text{m/s}^2 ][ W_{app} = 476.52 , text{N} ]The reading on the scale, which is the apparent weight of the passenger, is approximately 476.52 Newtons.Explanation of Apparent Weight:Apparent weight is the weight of an object as it is perceived in a non-inertial frame of reference, which in this case is the accelerating elevator. When the elevator accelerates downward, the passenger feels lighter because the scale (part of the elevator) is also accelerating downward. This means the normal force exerted by the scale on the passenger is reduced, resulting in a lower reading on the scale compared to if the elevator were at rest or moving at a constant velocity. Conversely, if the elevator were accelerating upward, the passenger would feel heavier, and the scale would show a higher reading because the normal force would be greater, compensating for the additional downward force due to the upward acceleration.

🔑:To determine if the driver of Car 1 was speeding, we'll use the principles of physics, specifically focusing on the kinetic energy and the work done by friction to stop Car 1 before the crash. We can calculate the initial speed of Car 1 based on the skid marks and then compare it to the speed limit. Given Values:- Mass of Car 1 ((m_1)) = 1875 kg- Mass of Car 2 ((m_2)) = 1051 kg (not directly relevant for calculating the speed of Car 1 before the crash)- Distance of Car 1's skid marks ((d)) = 16 meters- Speed limit = 50 km/h = 13.89 m/s (converted to meters per second for consistency) Calculations:1. Kinetic Energy and Work Done by Friction: The work done by friction ((W)) to stop Car 1 is equal to the change in kinetic energy ((Delta KE)) of Car 1. Since Car 1 comes to a complete stop, its final kinetic energy is 0. Thus, (W = Delta KE = KE_i - 0 = KE_i), where (KE_i) is the initial kinetic energy of Car 1.2. Work Done by Friction: The work done by friction can also be calculated using the formula (W = F_d cdot d), where (F_d) is the force of friction and (d) is the distance over which the force is applied. For kinetic friction, (F_d = mu_k cdot m cdot g), where (mu_k) is the coefficient of kinetic friction, (m) is the mass of the car, and (g) is the acceleration due to gravity (approximately 9.81 m/s^2).3. Equating Work and Kinetic Energy: We equate the work done by friction to the initial kinetic energy of Car 1: [ mu_k cdot m cdot g cdot d = frac{1}{2} cdot m cdot v_i^2 ] Solving for (v_i), the initial velocity (speed) of Car 1: [ v_i = sqrt{2 cdot mu_k cdot g cdot d} ]4. Coefficient of Kinetic Friction ((mu_k)): The coefficient of kinetic friction varies depending on the surfaces in contact. For tires on dry asphalt, a common value for (mu_k) is about 0.7. However, this value can range, and for the sake of this calculation, we'll use 0.7 as a reasonable estimate. Calculation with Given Values:Substituting the given values and (mu_k = 0.7) into the equation for (v_i):[v_i = sqrt{2 cdot 0.7 cdot 9.81 cdot 16}][v_i = sqrt{2 cdot 0.7 cdot 9.81 cdot 16} = sqrt{219.37}][v_i approx 14.81 , text{m/s}] Conversion to km/h:To compare with the speed limit, convert (v_i) from m/s to km/h:[v_i approx 14.81 , text{m/s} times frac{3600 , text{s/h}}{1000 , text{m/km}} approx 53.41 , text{km/h}] Conclusion:Given the calculations, the initial speed of Car 1 before the crash was approximately 53.41 km/h. This speed is slightly above the speed limit of 50 km/h, suggesting that the driver of Car 1 was likely speeding at the time of the collision. However, it's essential to note that this calculation involves several assumptions, including the coefficient of kinetic friction and the accuracy of the skid mark distance. In a real-world investigation, more precise data and potentially additional factors would be considered.