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Chapter 2: Problem 34

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

Expert verified
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

01

Write down the given displacement equation

The displacement \(x(t)\) of the moving object is given by the following equation: $$ x(t)=(4.35 \mathrm{~m})+(25.9 \mathrm{~m} / \mathrm{s}) t-\left(11.79 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} $$
02

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: $$ \frac{dx(t)}{dt} = \frac{d}{dt}\left((4.35 \mathrm{~m})+(25.9 \mathrm{~m} / \mathrm{s}) t-\left(11.79 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}\right) $$
03

Evaluate the derivative

Compute the derivative with respect to t for each term in the function, yielding: $$ \frac{dx(t)}{dt} = 0 + (25.9 \mathrm{~m} / \mathrm{s}) -2 \left(11.79 \mathrm{~m} / \mathrm{s}^{2}\right) t $$
04

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: $$ t = \frac{25.9 \mathrm{m/s}}{2 \cdot 11.79 \mathrm{m/s^2}} $$
05

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.
06

Calculate the maximum displacement

Plugging the value of t back into the original equation for x(t), we can find the maximum displacement: $$ x_{max} = (4.35 \mathrm{~m})+(25.9 \mathrm{~m} / \mathrm{s}) (1.099 \mathrm{s})-\left(11.79 \mathrm{~m} / \mathrm{s}^{2}\right) (1.099 \mathrm{s})^{2} $$
07

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.

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