Question

The position of an object as a function of time is given as \(x=A t^{3}+B t^{2}+C t+D .\) The constants are \(A=2.1 \mathrm{~m} / \mathrm{s}^{3}\) \(B=1.0 \mathrm{~m} / \mathrm{s}^{2}, C=-4.1 \mathrm{~m} / \mathrm{s},\) and \(D=3 \mathrm{~m}\) a) What is the velocity of the object at \(t=10.0 \mathrm{~s}\) ? b) At what time(s) is the object at rest? c) What is the acceleration of the object at \(t=0.50 \mathrm{~s} ?\) d) Plot the acceleration as a function of time for the time interval from \(t=-10.0 \mathrm{~s}\) to \(t=10.0 \mathrm{~s}\).

Short answer

Question: Calculate the object's velocity at t=10.0 s, find the times the object is at rest, the object's acceleration at t=0.50 s, and plot the acceleration as a function of time from t = -10.0 s to t=10.0 s, given the position function \(x(t) = 2.1t^3 + 1.0t^2 - 4.1t + 3\). Answer: The object's velocity at t=10.0 s is 645.9 m/s. The object is at rest at t ≈ 0.42 s and t ≈ -1.52 s. The object's acceleration at t=0.50 s is 8.3 m/s². To plot the acceleration as a function of time from t = -10.0 s to t = 10.0 s, use the function \(a(t) = 12.6t + 2.0\), resulting in a straight line with a positive slope, starting at \(a(-10.0) = -123.0 m/s^2\) and ending at \(a(10.0) = 127.0 m/s^2\).

Step-by-step solution

a) Calculate the object's velocity at t=10.0 s.

The velocity of an object is the derivative of its position function with respect to time. Given the position function: \(x(t) = 2.1t^3 + 1.0t^2 - 4.1t + 3\) Now, differentiate the position function with respect to time: \(v(t) = \dfrac{dx}{dt} = 6.3t^2 + 2.0t - 4.1\) To find the velocity at t=10.0 s, substitute this value of t into the velocity function: \(v(10.0) = 6.3(10.0)^2 + 2.0(10.0) - 4.1 = 6.3(100) + 20 - 4.1 = 630 + 15.9 = 645.9 m/s\) The object's velocity at t=10.0 s is 645.9 m/s.

b) Find the times the object is at rest.

An object is at rest when its velocity is zero. Using the velocity function we obtained previously, we set it to zero and solve for t. \(v(t) = 6.3t^2 + 2.0t - 4.1 = 0\) This is a quadratic equation, and we can use the quadratic formula to find the roots: \(t = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) In this case, a = 6.3, b = 2.0, and c = -4.1. Substituting these values, we get: \(t = \dfrac{-2\pm\sqrt{2^2 - 4(6.3)(-4.1)}}{2(6.3)}\) Now solve for t: \(t = \dfrac{-2\pm\sqrt{4+102.52}}{12.6} = \dfrac{-2\pm\sqrt{106.52}}{12.6}\) Calculating the two possible values of t: \(t_1 = \dfrac{-2+\sqrt{106.52}}{12.6} \approx 0.42 s\) \(t_2 = \dfrac{-2-\sqrt{106.52}}{12.6} \approx -1.52 s\) The object is at rest at t ≈ 0.42 s and t ≈ -1.52 s.

c) Find the object's acceleration at t=0.50 s.

The acceleration of the object is the derivative of its velocity function with respect to time. We differentiate the velocity function we obtained in part (a) with respect to time: \(a(t) = \dfrac{dv}{dt} = 12.6t + 2.0\) Now, we substitute t=0.50 s to find the acceleration at this time: \(a(0.50) = 12.6(0.50) + 2.0 = 6.3 + 2.0 = 8.3 m/s^2\) The object's acceleration at t=0.50 s is 8.3 m/s².

d) Plot the acceleration as a function of time.

We have obtained the acceleration function in part (c): \(a(t) = 12.6t + 2.0\). Now, we can create a plot of this function in the given time interval from t = -10.0 s to t = 10.0 s. Use your preferred graphing software or graphing calculator to plot the function \(a(t) = 12.6t + 2.0\) in the interval \(-10.0 \leq t \leq 10.0\). The resulting graph is a straight line with a positive slope, starting at \(a(-10.0) = -123.0 m/s^2\) and ending at \(a(10.0) = 127.0 m/s^2\)

Key concepts

Understanding the Velocity-Time Relationship

The velocity-time relationship is pivotal in kinematics, which depicts how an object's velocity changes over time. In our exercise, the object's position function was given, and we found its velocity function through differentiation. Fundamentally, the slope of the position-time graph at any instant represents the velocity at that moment.

For instance, the equation derived, \(v(t) = 6.3t^2 + 2.0t - 4.1\), encapsulates how the object's velocity varies with time. By plugging in different values of \(t\), we can predict the velocity at that specific time. This relationship is essential for predicting future motion and analyzing past motion of objects.

Decoding the Acceleration-Time Graph

An acceleration-time graph provides us with insights into how an object’s acceleration changes as time progresses. Acceleration being the derivative of velocity with respect to time implies that a graph plotting this relationship is simply a visual representation of the object's rate of change of velocity.

In the given exercise, when we plotted the acceleration function \(a(t) = 12.6t + 2.0\), it created a straight line with the slope representing constant change in acceleration. From this graph, we can discern the nature of the object's motion, whether it's uniformly accelerating, decelerating, or undergoing complex dynamic changes.

Applying Motion Equations

Motion equations, also known as equations of motion, are formulas that describe the movement of an object. They combine position, velocity, acceleration, and time to provide a complete description of the motion. In classical mechanics, there are generally four kinematic equations used to solve motion problems when the acceleration is constant.

In this problem, however, our object does not have a constant acceleration, which is evident from the acceleration function \(a(t) = 12.6t + 2.0\). Therefore, we do not use the standard kinematic equations but instead apply calculus to derive the relationships between velocity, position, and acceleration directly from the position-time function.

The Role of Differentiation in Physics

Differentiation is a mathematical process used extensively in physics to understand the way quantities change. It allows us to calculate velocity from position, and acceleration from velocity by finding the rate at which these quantities change with time.

In the given problem, differentiation was utilized to determine the velocity and acceleration functions from the position function. It involved taking the first derivative to get the velocity \(v(t)\), and the second derivative to obtain acceleration \(a(t)\). This concept of differentiation enables physicists and engineers to predict and analyze motion in a wide array of applications.