Answer The Questions Below: Show All Your Work And Formulas

Answer The Questions Below Show All Your Work And Formulas

The assignment involves analyzing motion scenarios through graphs, calculations of displacement, velocity, acceleration, and understanding the effects of forces such as gravity and friction. The questions require applying kinematic equations, understanding the nature of acceleration, and interpreting motion graphs, as well as performing algebraic calculations to determine distances and stopping points for objects on inclined planes and under deceleration.

Paper For Above instruction

Introduction

Understanding motion requires analyzing various physical parameters such as position, velocity, acceleration, and the forces acting on objects. These parameters are interconnected through fundamental kinematic equations, which describe how objects move under different conditions. The following analysis addresses specific scenarios involving motion on inclined planes and deceleration, employing graphical and algebraic methods to interpret and predict motion characteristics.

Question 1: Analyzing Graphs of Position and Acceleration

Suppose we are given a velocity-time or acceleration-time graph (as mentioned in the prompt). The objective is to determine when the object has positive or negative acceleration based on the graph.

• Positive acceleration: The object exhibits positive acceleration when the acceleration curve is above the time axis, indicating that the acceleration vector points in the same direction as the positive coordinate axis, typically causing the velocity to increase.
• Negative acceleration: Conversely, negative acceleration occurs when the acceleration curve is below the time axis, resulting in a decrease in velocity, often referred to as deceleration.

The core concept is that the sign of the acceleration directly influences whether the object's velocity increases or decreases. Graphical analysis involves observing where the acceleration curve crosses the zero line and noting its positive or negative regions. To visualize these phenomena, plotting the position and acceleration curves over time would reveal how the object's displacement and acceleration evolve, assuming it starts from zero displacement at t=0.

Question 2: Motion of a Frictionless Cart on an Incline

A cart released from rest on a frictionless inclined plane inclined at 30° demonstrates acceleration due to gravity components along the incline. To calculate the distance traveled in 3 seconds, consider the following:

• Acceleration along the incline: a = g * sin(θ), where g = 9.8 m/s² and θ = 30°.
• Calculation of acceleration: a = 9.8 sin(30°) = 9.8 0.5 = 4.9 m/s².
• Using kinematic equations: For an initial velocity v₀=0, the displacement after time t is s = v₀ t + ½ a * t².
• Distance: s = 0 + ½ 4.9 (3)² = 0.5 4.9 9 = 22.05 meters.

Graphically, the position-time graph is a parabola opening upwards, showing increasing displacement over time. The velocity-time graph would be a straight line starting at zero with slope equal to the acceleration (4.9 m/s²). The acceleration graph remains constant at 4.9 m/s² due to the constant acceleration component along the incline.

Question 3: Deceleration and Complete Stop After 5 Seconds

Initially, the cart moves under the influence of gravity and then the brake applies a deceleration of 6.9 m/s² when off the incline to bring it to a halt. The key steps involve calculating the distance traveled during braking and total displacement, including the slide on the incline prior to braking.

• Velocity at 5 seconds: For the initial phase (before brakes), velocity v = a t, with a = 4.9 m/s². After 5 seconds, v = 4.9 5 = 24.5 m/s.
• Braking distance: Using v² = v₀² + 2 a d, with final velocity v_f=0 and initial v₀=24.5 m/s, and acceleration a = -6.9 m/s² (negative for deceleration), solve for d:

0 = (24.5)² + 2 (-6.9) d → d = (24.5)² / (2 * 6.9) ≈ 600.25 / 13.8 ≈ 43.5 meters.

• Total distance traveled: Sum of the distance traveled in the first 5 seconds along the incline plus the braking distance.
• Distance traveled on inclined plane before braking: s = ½ a t² = ½ 4.9 (5)² = 0.5 4.9 25 = 61.25 meters.
• Adding the braking distance: total = 61.25 + 43.5 = 104.75 meters.

Graphically, the position curve initially rises as the cart accelerates down the incline and then transitions into a downward curve as braking forces slow and stop the cart. The velocity-time graph peaks at the 5-second mark and then decreases to zero during deceleration. The acceleration graph shows a steady positive value during acceleration and a constant negative value during braking.

Conclusion

Analyzing the motion scenarios involving inclined planes and deceleration illustrates the application of fundamental physics principles. Graphical interpretations of position, velocity, and acceleration help visualize how objects move under such influences, while algebraic calculations enable precise determination of distances traveled and stopping points. Mastery of these concepts is essential in understanding kinematic behavior in real-world systems, from simple inclined planes to complex braking mechanisms.

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