Determining Velocity And Motion Of A Particle In Physics Problems And Solutions

In the realm of physics, understanding the motion of objects is paramount. This article delves into two specific problems related to particle motion, providing a comprehensive analysis and solution for each. The first problem focuses on determining the velocity of a particle given its position as a function of time. The second problem explores the motion of a particle defined by a more complex relation, requiring us to find when the velocity is zero, as well as the velocity, acceleration, and total distance traveled at a specific time. By meticulously examining these problems, we aim to enhance your grasp of kinematic concepts and problem-solving techniques.

Problem vi Determining Velocity from Position Function

Velocity from Position is a fundamental concept in kinematics, and this problem provides a practical application. Given the position of a particle as a function of time, x = 2t³, where t is in seconds and x is in meters, our objective is to determine the velocity, v, when t = 2 seconds. To solve this, we need to recall the relationship between position and velocity: velocity is the time derivative of position. In mathematical terms, v = dx/dt. This means we need to differentiate the given position function with respect to time.

Applying the power rule of differentiation to x = 2t³, we get:

dx/dt = 6t²

This equation represents the instantaneous velocity of the particle at any time t. Now, to find the velocity at t = 2 seconds, we simply substitute t = 2 into the equation:

v = 6(2)² = 6 * 4 = 24 m/s

Therefore, the velocity of the particle at t = 2 seconds is 24 m/s. This result signifies the rate at which the particle's position is changing at that specific moment. The positive sign indicates that the particle is moving in the positive direction along the x-axis. Understanding this process of finding velocity from position is crucial for analyzing various motion scenarios in physics. The derivative, in this context, gives us the instantaneous rate of change, which is a powerful tool in understanding dynamic systems.

This problem highlights the importance of calculus in physics. The derivative provides a precise way to determine velocity, which is essential for predicting and understanding the motion of objects. Furthermore, this concept extends to more complex scenarios where the position function might involve higher-order polynomials or trigonometric functions. The underlying principle, however, remains the same: differentiate the position function to obtain the velocity function.

In conclusion, determining velocity from a position function is a core skill in physics. By understanding the relationship between position and velocity, and by applying the principles of calculus, we can accurately describe and predict the motion of particles. This problem serves as a foundational example for more advanced topics in kinematics and dynamics.

Problem vii Analyzing Motion with a Complex Position Function

This problem presents a more intricate scenario, requiring a comprehensive analysis of a particle's motion. The motion of the particle is defined by the relation x = t³ - 6t² - 36t - 40, where x is the position and t is the time. Our task is to determine several key aspects of the motion: (a) when the velocity is zero, (b) the velocity, the acceleration, and the total distance traveled when t = 4 s. This problem necessitates the application of both differentiation and a careful consideration of the particle's changing direction.

(a) When is the Velocity Zero?

To find when the velocity is zero, we first need to determine the velocity function. As before, velocity is the time derivative of position:

v = dx/dt = 3t² - 12t - 36

Now, we set the velocity function equal to zero and solve for t:

3t² - 12t - 36 = 0

Dividing the equation by 3 simplifies it:

t² - 4t - 12 = 0

This is a quadratic equation that can be factored:

(t - 6)(t + 2) = 0

This gives us two possible solutions for t: t = 6 s and t = -2 s. Since time cannot be negative, we discard t = -2 s. Therefore, the velocity of the particle is zero at t = 6 seconds. This indicates a point where the particle momentarily stops before potentially changing direction.

(b) Velocity, Acceleration, and Total Distance Traveled at t = 4 s

First, let's find the velocity at t = 4 s. We already have the velocity function:

v = 3t² - 12t - 36

Substituting t = 4:

v = 3(4)² - 12(4) - 36 = 48 - 48 - 36 = -36 m/s

This tells us that at t = 4 seconds, the particle is moving with a velocity of -36 m/s. The negative sign indicates that the particle is moving in the negative direction along the x-axis.

Next, we need to find the acceleration. Acceleration is the time derivative of velocity:

a = dv/dt = 6t - 12

Substituting t = 4:

a = 6(4) - 12 = 24 - 12 = 12 m/s²

The acceleration at t = 4 seconds is 12 m/s². This positive value indicates that the particle is accelerating in the positive direction, even though its velocity is negative. This means the particle is slowing down as it moves in the negative direction.

Finally, we need to determine the total distance traveled. This is not simply the position at t = 4 seconds because the particle might have changed direction. To find the total distance, we need to consider the time interval where the velocity changes sign. We know the velocity is zero at t = 6 seconds. So, we need to analyze the motion between t = 0 and t = 4 seconds.

At t = 0, the position is:

x(0) = (0)³ - 6(0)² - 36(0) - 40 = -40 m

At t = 4, the position is:

x(4) = (4)³ - 6(4)² - 36(4) - 40 = 64 - 96 - 144 - 40 = -216 m

Since the velocity is negative between t=0 and t=4, the particle moves from x=-40 m to x=-216 m. The distance traveled in this interval is:

|Distance| = |-216 - (-40)| = |-176| = 176 m

Therefore, the total distance traveled by the particle at t = 4 seconds is 176 meters. This calculation takes into account the direction of motion and ensures that we are measuring the actual path length traveled by the particle.

In summary, at t = 4 seconds, the velocity is -36 m/s, the acceleration is 12 m/s², and the total distance traveled is 176 meters. This comprehensive analysis demonstrates the power of calculus in understanding the nuances of particle motion, including changes in direction and the distinction between displacement and distance.

These two problems illustrate the fundamental principles of kinematics and the application of calculus in solving motion-related problems. By understanding the relationship between position, velocity, and acceleration, and by mastering the techniques of differentiation, we can effectively analyze and predict the motion of objects. The first problem highlighted the direct application of the derivative to find velocity from a position function, while the second problem showcased a more complex scenario requiring a deeper understanding of motion analysis, including finding when velocity is zero and calculating total distance traveled. These concepts are crucial for further studies in physics and engineering.