Body Motion Problems And Solutions S(t)

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This article explores problems related to body motion described by the function S(t), focusing on finding key parameters such as maximum acceleration and velocity. We will analyze two specific problems: one involving a polynomial function for displacement and the other involving a square root function. These problems are fundamental in understanding the principles of kinematics and dynamics in physics.

1. Problem: Finding the Time of Maximum Acceleration

Keywords: Body Motion, Acceleration, Maximum Acceleration, Time, Kinematics, Polynomial Function

Original Question: Тело движется по закону S(t) = -1/12t^4 + 2/3t^3 - 3/2t^2. Пройдя через сколько секунд от начала движения, ускорение тела будет наибольшим?

Rewritten Question: A body moves according to the law S(t) = -1/12t^4 + 2/3t^3 - 3/2t^2. After how many seconds from the beginning of the motion will the body's acceleration be the greatest?

Solution Approach:

To solve this problem, we need to understand the relationship between displacement, velocity, and acceleration. The velocity, v(t), is the first derivative of the displacement function, S(t), with respect to time, t. The acceleration, a(t), is the first derivative of the velocity function or the second derivative of the displacement function.

  1. Find the velocity function:

    v(t) = dS(t)/dt

    Given S(t) = -1/12t^4 + 2/3t^3 - 3/2t^2, we differentiate with respect to t:

    v(t) = -1/3t^3 + 2t^2 - 3t

  2. Find the acceleration function:

    a(t) = dv(t)/dt = d2S(t)/dt2

    Differentiating v(t) with respect to t:

    a(t) = -t^2 + 4t - 3

  3. Find the time at which acceleration is maximum:

    To find the maximum acceleration, we need to find the critical points of the acceleration function. This is done by finding where the derivative of the acceleration function, a'(t), is equal to zero or undefined. In this case, a(t) is a quadratic function, so its maximum (or minimum) occurs at its vertex.

    a'(t) = da(t)/dt

    Differentiating a(t) = -t^2 + 4t - 3 with respect to t:

    a'(t) = -2t + 4

    Setting a'(t) = 0 to find critical points:

    -2t + 4 = 0

    2t = 4

    t = 2

  4. Verify that it's a maximum:

    To verify that t = 2 corresponds to a maximum, we can use the second derivative test. The second derivative of the acceleration function, a''(t), is:

    a''(t) = d2a(t)/dt2

    Differentiating a'(t) = -2t + 4 with respect to t:

    a''(t) = -2

    Since a''(t) = -2 is negative, the acceleration function has a maximum at t = 2.

  5. Compare with the given options:

    The options are:

    • A) 2.5
    • B) 1.5
    • C) 3
    • D) 1.75

    The correct answer is not directly among the options, which suggests a potential error in the original options or question. However, based on our calculations, the time at which the acceleration is maximum is t = 2 seconds.

Conclusion for Problem 1:

In summary, to find the time at which the acceleration of the body is greatest, we calculated the first and second derivatives of the displacement function to obtain the velocity and acceleration functions, respectively. We then found the critical points of the acceleration function by setting its derivative equal to zero. The second derivative test confirmed that the critical point corresponds to a maximum. Therefore, the acceleration is greatest at t = 2 seconds. However, since this value isn't in the provided options, it indicates a discrepancy.

2. Problem: Finding Velocity at a Specific Time

Keywords: Body Motion, Velocity, Time, Kinematics, Square Root Function

Original Question: Тело движется по закону S(t) = t√t. Найдите...

Rewritten Question: A body moves according to the law S(t) = t√t. Find...

To make this problem complete, we need to specify what we want to find. Let's assume we want to find the velocity at a specific time, say t = 4 seconds.

Completed Question: A body moves according to the law S(t) = t√t. Find the velocity of the body at t = 4 seconds.

Solution Approach:

Similar to the previous problem, we need to find the velocity function by differentiating the displacement function with respect to time.

  1. Rewrite the displacement function:

    S(t) = t√t can be rewritten using exponents as: S(t) = t * t^(1/2) = t^(3/2)

  2. Find the velocity function:

    v(t) = dS(t)/dt

    Differentiating S(t) = t^(3/2) with respect to t:

    v(t) = (3/2)t^(1/2)

  3. Evaluate the velocity at t = 4 seconds:

    Substitute t = 4 into the velocity function:

    v(4) = (3/2)(4)^(1/2)

    v(4) = (3/2) * 2

    v(4) = 3

Conclusion for Problem 2:

In this problem, we found the velocity of the body at a specific time by first rewriting the displacement function using exponents and then differentiating it to obtain the velocity function. By substituting t = 4 seconds into the velocity function, we found that the velocity of the body at that time is 3 units per second.

General Strategies for Solving Body Motion Problems

Keywords: Problem-Solving Strategies, Body Motion, Kinematics, Derivatives, Displacement, Velocity, Acceleration

To effectively solve body motion problems, it's essential to understand the fundamental relationships between displacement, velocity, and acceleration. These quantities are interconnected through calculus, where velocity is the derivative of displacement with respect to time, and acceleration is the derivative of velocity (or the second derivative of displacement) with respect to time.

Key Steps in Solving Body Motion Problems:

  1. Understand the given information: Carefully read the problem statement and identify the given quantities and the quantities you need to find. This often involves recognizing the displacement function S(t) and the specific questions related to velocity v(t) or acceleration a(t).

  2. Apply relevant formulas: The core formulas in kinematics are derived from calculus:

    • Velocity (v(t)) = dS(t)/dt: This formula calculates the instantaneous velocity by finding the derivative of the displacement function with respect to time.
    • Acceleration (a(t)) = dv(t)/dt = d2S(t)/dt2: This formula calculates the instantaneous acceleration by finding the first derivative of the velocity function or the second derivative of the displacement function with respect to time.

  3. Calculate derivatives: Use the rules of differentiation to find the velocity and acceleration functions. Common rules include the power rule, product rule, quotient rule, and chain rule. For polynomial functions like in the first problem, the power rule is frequently used. For functions involving square roots, rewriting the function using fractional exponents simplifies differentiation.

  4. Find critical points: To determine maximum or minimum values (e.g., maximum acceleration), find the critical points of the relevant function (e.g., acceleration function). Critical points occur where the derivative of the function is zero or undefined. Setting the first derivative to zero and solving for t identifies these points.

  5. Use the second derivative test: To verify whether a critical point corresponds to a maximum or minimum, apply the second derivative test. If the second derivative at the critical point is positive, the function has a local minimum. If it's negative, the function has a local maximum. If it's zero, the test is inconclusive, and further analysis is needed.

  6. Evaluate at specific times: If the problem asks for the velocity or acceleration at a particular time, substitute that time into the respective function. For example, to find the velocity at t = 4 seconds, substitute t = 4 into the velocity function v(t).

  7. Interpret the results: Ensure that the results are physically meaningful. For instance, a negative velocity indicates movement in the opposite direction. Check the units and ensure they are consistent throughout the problem.

Advanced Problem-Solving Techniques:

  1. Understanding Motion Graphs: Visualizing motion through graphs can provide additional insights. Plotting displacement, velocity, and acceleration as functions of time helps to understand the motion's characteristics, such as changes in speed, direction, and the presence of constant or varying acceleration.

  2. Dealing with Non-Constant Acceleration: Problems involving non-constant acceleration may require more advanced techniques, such as integrating the acceleration function to find velocity or using numerical methods if an analytical solution is not possible.

  3. Using Vector Calculus: For motion in two or three dimensions, vector calculus is necessary. Displacement, velocity, and acceleration become vector quantities, and the principles of calculus are applied component-wise.

  4. Incorporating Initial Conditions: Many problems provide initial conditions (e.g., initial position and velocity). These conditions are crucial for determining the constants of integration when integrating acceleration to find velocity and integrating velocity to find displacement.

Common Mistakes to Avoid:

  1. Incorrect Differentiation: Ensure the differentiation rules are applied correctly. A common mistake is mishandling fractional exponents or applying the power rule incorrectly.

  2. Forgetting Units: Always include units in the final answer. Incorrect units indicate a fundamental misunderstanding of the problem.

  3. Misinterpreting Maxima and Minima: Ensure that the second derivative test is used correctly to distinguish between maxima and minima.

  4. Ignoring Initial Conditions: Initial conditions are crucial for finding specific solutions. Neglecting them can lead to a general solution rather than the particular solution required by the problem.

  5. Algebraic Errors: Carefully review each step to avoid algebraic errors, which can easily lead to incorrect results.

By following these strategies and avoiding common mistakes, you can effectively solve a wide range of body motion problems, enhancing your understanding of kinematics and dynamics.