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Refer to the following. Suppose a ball is thrown straight upward with an initial velocity (that is, velocity at the time of release) of \(128 \mathrm{ft} / \mathrm{sec}\) and that the point at which the ball is released is considered to be at zero height. Then the height \(s(t)\) in feet of the ball at time \(t\) in seconds is given by \(s(t)=-16 t^{2}+128 t .\) Let \(v(t)\) be the instantaneous velocity at time \(t\). Find \(v(t)\) for the indicated values of \(t\). \(v(6)\)

Short Answer

Expert verified
The instantaneous velocity \( v(6) = -64 \mathrm{ft/sec} \).

Step by step solution

01

Understanding Instantaneous Velocity

Instantaneous velocity is the derivative of the position function with respect to time. In our equation, the position function is given by \( s(t) = -16t^2 + 128t \). Therefore, the instantaneous velocity function \( v(t) \) can be found by differentiating \( s(t) \) with respect to \( t \).
02

Calculating the Derivative

Differentiate the height function \( s(t) = -16t^2 + 128t \). Using basic differentiation rules, the derivative of \(-16t^2\) is \(-32t\) and the derivative of \(128t\) is \(128\). Therefore, the velocity function \( v(t) \) is \( v(t) = -32t + 128 \).
03

Substitute the Given Time Value

To find \( v(6) \), substitute \( t = 6 \) into the velocity function. This gives us: \[ v(6) = -32(6) + 128 \]
04

Performing Calculations

Calculate \(-32(6) + 128\):- First, calculate \(-32 \times 6 = -192\).- Then, add \(128\) to \(-192\) to get \(-192 + 128 = -64\).
05

Conclusion

The value of the instantaneous velocity \( v(6) \) is \(-64 \mathrm{ft/sec}\). This means that at \(6\) seconds, the ball is moving downward at a speed of \(64\) feet per second.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Instantaneous Velocity
Instantaneous velocity is a measure of how fast an object is moving at a specific moment. Imagine you are driving a car. The speedometer gives you the instantaneous speed: your speed at that exact moment. When dealing with a position function like our ball's height function, the instantaneous velocity is found by determining how fast the height is changing at any given second. In mathematical terms, this is represented as the derivative of the position function, which tells us the rate of change of position with respect to time. Calculating the instantaneous velocity at a certain moment involves substituting that moment's time into the velocity function derived from differentiation. Thus, it provides valuable insight into the speed and direction of a moving object, like a ball thrown in the air.
Differentiation
Differentiation in calculus is the process of finding the derivative of a function. The derivative measures how a function changes as its input changes. To understand it better, think about how a car's speedometer changes as you accelerate. Differentiation helps us determine the rate of change, similar to how you feel the press on your back as the car speeds up. In our problem, we differentiated the position function of the ball to find the velocity function. This is done by applying rules like the power rule, where the derivative of \(-16t^2\) becomes \-32t\ and \(128t\) becomes \128\. Differentiation helps us convert position functions into velocity functions, providing a powerful tool for understanding dynamics in physical systems.
Position Function
A position function describes the location of an object at a particular time. In physics, it's often expressed as a function of time and gives us details like the height, distance, or location of a moving object.For the ball in our problem, the position function is \(s(t) = -16t^2 + 128t\). This function balances effects such as gravitational pull (represented by the \(-16t^2\) term) and the initial velocity of the ball (represented by the \(128t\) term). By evaluating this function at any given time \(t\), we can determine the height of the ball at that particular moment. Understanding the position function is essential for predicting how the object moves over time and helps set the foundation for deriving other important concepts like velocity and acceleration.

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Most popular questions from this chapter

Cost Function A steel plant has a cost function \(C(x)\), where \(x\) is measured in tons of steel and \(C\) is measured in dollars. Suppose that when \(x=150,\) the instantaneous rate of change of cost with respect to tons is \(300 .\) Explain what this means.

Find the limits graphically. Then confirm algebraically. $$ \lim _{x \rightarrow 1} \frac{x^{2}-1}{x-1} $$

On your computer or graphing calculator, graph \(y=\) \(f(x)=\sin x\) in radian mode using a window with dimensions [-6.14,6.14] by \([-1,1] .\) As you see, this function moves back and forth between -1 and \(1 .\) We wish to estimate \(f^{\prime}(0)\). For this purpose, graph using a window with dimensions [-0.5,0.5] by \([-0.5,0.5] .\) From the graph, estimate \(f^{\prime}(0)\).

Let \(R(x)\) be a revenue function for a firm, with \(x\) the number of units sold and \(R(x)\) the total revenue in dollars from selling \(x\) units. If \(R^{\prime}(1000)=6,\) what is the approximate revenue from the sale of the 1001 st unit?

Cost Curve Johnston \(^{68}\) made a statistical estimation of the cost-output relationship for 40 firms. The data for one of the firms is given in the following table. $$ \begin{array}{|c|cccccc|} \hline x & 2.5 & 3 & 4 & 5 & 6 & 7 \\ \hline y & 20 & 26 & 20 & 27 & 30 & 29 \\ \hline x & 11 & 22.5 & 29 & 37 & 43 & \\ \hline y & 42 & 53 & 75 & 73 & 78 & \\ \hline \end{array} $$ Here \(x\) is the output in millions of units, and \(y\) is the total cost in thousands of pounds sterling. a. Determine both the best-fitting line using least squares and the square of the correlation coefficient. Graph. b. Determine both the best-fitting quadratic using least squares and the square of the correlation coefficient. Graph. c. Which curve is better? Why? d. Using the quadratic cost function found in part (b) and the approximate derivative found on your computer or graphing calculator, graph the marginal cost. What is happening to marginal cost as output increases? Explain what this means.

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