Question

The trajectory of an object is given by the equation $$ x(t)=(4.35 \mathrm{~m})+(25.9 \mathrm{~m} / \mathrm{s}) t-\left(11.79 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} $$ a) For which time \(t\) is the displacement \(x(t)\) at its maximum? b) What is this maximum value?

Short answer

Answer: a) The time t when the displacement x(t) is at its maximum is approximately 1.099 s. b) The maximum value of the displacement x(t) is approximately 16.72 m.

Step-by-step solution

Write down the given displacement equation

The displacement \(x(t)\) of the moving object is given by the following equation: $$ x(t)=(4.35 \mathrm{~m})+(25.9 \mathrm{~m} / \mathrm{s}) t-\left(11.79 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} $$

Compute the derivative of the function

To find the maximum value of x(t), we have to first find the critical points of the function. We can determine these critical points by calculating the first derivative of the function with respect to time t and setting it equal to zero. Thus, compute the derivative of the function as follows: $$ \frac{dx(t)}{dt} = \frac{d}{dt}\left((4.35 \mathrm{~m})+(25.9 \mathrm{~m} / \mathrm{s}) t-\left(11.79 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}\right) $$

Evaluate the derivative

Compute the derivative with respect to t for each term in the function, yielding: $$ \frac{dx(t)}{dt} = 0 + (25.9 \mathrm{~m} / \mathrm{s}) -2 \left(11.79 \mathrm{~m} / \mathrm{s}^{2}\right) t $$

Find the critical points

To find the critical points, we must set the derivative of the function equal to zero and solve for t: $$ (25.9 \mathrm{~m} / \mathrm{s}) - 2 \left(11.79 \mathrm{~m} / \mathrm{s}^{2}\right) t = 0 $$ Now, solve for t: $$ t = \frac{25.9 \mathrm{m/s}}{2 \cdot 11.79 \mathrm{m/s^2}} $$

Calculate the time t for maximum displacement

Compute the value of t: $$ t = \frac{25.9 \mathrm{m/s}}{2 \cdot 11.79 \mathrm{m/s^2}} \approx 1.099 \mathrm{s} $$ The displacement x(t) is at its maximum when t ≈ 1.099s.

Calculate the maximum displacement

Plugging the value of t back into the original equation for x(t), we can find the maximum displacement: $$ x_{max} = (4.35 \mathrm{~m})+(25.9 \mathrm{~m} / \mathrm{s}) (1.099 \mathrm{s})-\left(11.79 \mathrm{~m} / \mathrm{s}^{2}\right) (1.099 \mathrm{s})^{2} $$

Evaluate the maximum displacement

Compute the maximum displacement x(t), yielding: $$ x_{max} \approx 16.72 \mathrm{~m} $$ a) The time t when the displacement x(t) is at its maximum is approximately 1.099 s. b) The maximum value of the displacement x(t) is approximately 16.72 m.

Key concepts

Displacement

Displacement is a key concept in kinematics, a branch of physics that studies motion. Essentially, displacement refers to how much an object has moved from its original position. It is a vector quantity, which means it has both a magnitude and a direction. Imagine pushing a toy car across a table from the start to the finish line - the distance it covers in a straight line represents its displacement, not the path it took.

• Displacement considers only the initial and final positions.
• It is different from distance which measures the entire path taken.
• Typical units for displacement are meters (m), as shown in this exercise.

In the given exercise, displacement is described by the function:\[x(t) = (4.35 \text{ m}) + (25.9 \text{ m/s}) t - (11.79 \text{ m/s}^2) t^2\]This function relates displacement \(x\) to time \(t\), considering factors like initial position, velocity, and acceleration.

Derivative

The derivative is a fundamental tool in calculus used to understand how a quantity changes. In the context of motion, it can tell us how the displacement of an object changes with time. Calculating the derivative of the displacement function helps us understand the object's velocity. This is crucial because it can indicate when the displacement reaches a pivotal point, such as a maximum.

• The derivative \(\frac{dx(t)}{dt}\) represents the rate of change of displacement over time, which is essentially velocity.
• To find the derivative of \(x(t) = (4.35 \text{ m}) + (25.9 \text{ m/s}) t - (11.79 \text{ m/s}^2) t^2\), differentiate each term:
  \[\frac{dx(t)}{dt} = 0 + 25.9 \text{ m/s} - 2 \times 11.79 \text{ m/s}^2 \times t\]
• The derivative helps locate critical points where velocity is zero and displacement might be at a maximum or minimum.

Critical Points

Critical points are essential when analyzing functions like the displacement equation given in this exercise. They occur where the first derivative equals zero, indicating potential maximum or minimum values of the function. For motion, these points can show where an object changes its state of motion, such as when reaching maximum displacement.

• To find critical points, solve \(\frac{dx(t)}{dt} = 0\) for time \(t\).
• In our exercise, the equation:
  \[25.9 \text{ m/s} - 2 \times 11.79 \text{ m/s}^2 \times t = 0\]
  is solved to find \(t\).
• The solution, \(t \approx 1.099\text{ s}\), marks the critical point where displacement is potentially maximized.

By identifying these points, we can determine when the object is moving fastest or is at rest momentarily. This helps us calculate precise maximum or minimum displacement values.

Maximum Displacement

The maximum displacement refers to the highest value of displacement reached during the object's motion. Once the critical points are found, we can determine if these points result in a maximum. In this exercise, after calculating the critical point, the next step is to find out if this point represents the maximum displacement.

• To confirm maximum displacement, substitute the critical point time back into the original displacement equation.
• For \(t \approx 1.099\text{ s}\), substitute in the equation:
  \[x_{max} = 4.35 \text{ m} + 25.9 \text{ m/s} \times 1.099\text{ s} - 11.79 \text{ m/s}^2 \times (1.099\text{ s})^2\]
• Calculating this yields \(x_{max} \approx 16.72 \text{ m}\).

Thus, the object reaches its maximum displacement of approximately 16.72 meters at 1.099 seconds. Understanding maximum displacement is crucial for applications needing peak performance insights, such as in sports or engineering challenges.