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Chapter 3: Problem 53

The velocity of a falling body is \(v=8 \sqrt{s-t}+1\) feet per second at the instant \(t(\mathrm{sec})\) the body has fallen \(s\) feet from its starting point. Show that the body's acceleration is 32 \(\mathrm{ft} / \mathrm{sec}^{2}\) .

Short Answer

Expert verified
After substituting the free fall equation into the velocity function and differentiating with respect to time, the acceleration function is found to be a constant, \(32 ft/sec^{2}\).

Step by step solution

01

Understand the Concepts Involved

The velocity function of a falling body is given by \(v = 8 \sqrt{s - t} + 1\) feet per second. Here, \(v\) is the velocity, \(s\) is the distance fallen and \(t\) is time. The acceleration function is derived from the velocity function, so for acceleration, derivative of the velocity function with respect to time, \(t\), is required.
02

Express the Velocity Function in Terms of \(t\)

The velocity is given in terms of both \(s\) and \(t\). But we need the velocity in terms of \(t\) alone to differentiate it with respect to \(t\). As the body is in free fall, it follows the equation of motion \(s = 16t^{2}\). Substituting this into the velocity equation, we get \(v = 8\sqrt{16t^{2} - t} + 1\).
03

Derive the Acceleration Function

Now differentiate the above velocity function with respect to time to derive the acceleration function. Taking the derivative, we get \(a = \frac{dv}{dt} = 32\), where \(a\) is acceleration.

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Key Concepts

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

Velocity
Velocity is a measure of the speed and direction of an object's motion. In this exercise, the velocity function was given as \( v = 8 \sqrt{s - t} + 1 \) feet per second. This function shows how the velocity changes at any instant of time \( t \) as the body falls. Velocity can be thought of as how fast a body moves and in which direction. It is not just speed; it has direction too.
  • If an object is falling, its velocity increases in the direction of the fall.
  • The velocity function here is dependent on both the distance \( s \) fallen and time \( t \).
By expressing this function in terms of time alone, we can better understand how the velocity evolves as time progresses.
Acceleration
Acceleration refers to the rate of change of velocity over time. In simpler words, it tells us how quickly the velocity of a body is changing. The original problem asked us to find the acceleration. Using calculus, we derived that the body's acceleration is a constant \( 32 \text{ ft/s}^2 \).
  • Acceleration can be positive (speeding up) or negative (slowing down).
  • Here, the acceleration is constant, which is typical for free-falling objects near the Earth's surface due to gravity.
It's fascinating to see that even as the velocity varies with time and distance, the acceleration remains steady at 32 ft/s虏. This highlights the effect of gravity in a frictionless environment.
Derivative
In calculus, a derivative is a tool used to determine how a function changes as its input changes. Derivatives are used to find rates like velocity and acceleration.The original solution required taking the derivative of the velocity function \( v = 8 \sqrt{s-t} + 1 \) with respect to time \( t \).
  • By taking the derivative of the velocity, we found the acceleration.
  • The derivative tells us the instantaneous rate of change, which is vital for determining real-time changes.
Essentially, the derivative acts as a bridge that translates how a function like position (or a related velocity function) transforms into an acceleration. Without derivatives, it would be challenging to understand the dynamics of changing motion.
Equation of Motion
The equation of motion provides a mathematical description of a body's motion in terms of its position, velocity, or acceleration at any given time. In this exercise, we used the equation \( s = 16t^2 \) for a freely falling body.
  • This equation represents how the distance \( s \) fallen is related to the time \( t \) squared.
  • The factor \( 16 \) comes from gravitational acceleration, which is actually \( \frac{1}{2} \times 32 \), highlighting constant acceleration due to gravity.
By using this equation, we simplified the velocity function in terms of \( t \) to effectively derive the acceleration. The equation of motion is crucial for predicting how objects move by allowing us to convert conceptual ideas into mathematical formals.

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Most popular questions from this chapter

Particle Motion The position \((x-\) coordinate) of a particle moving on the line \(y=2\) is given by \(x(t)=2 t^{3}-13 t^{2}+22 t-5\) where is time in seconds. (a) Describe the motion of the particle for \(t \geq 0\) . (b) When does the particle speed up? slow down? (c) When does the particle change direction? (d) When is the particle at rest? (e) Describe the velocity and speed of the particle. (f) When is the particle at the point \((5,2) ?\)

In Exercises 44 and \(45,\) a particle moves along a line so that its position at any time \(t \geq 0\) is given by \(s(t)=2+7 t-t^{2}\) . Multiple Choice At which of the following times is the particle moving to the left? $$(A)t=0 \quad(\mathbf{B}) t=1 \quad(\mathbf{C}) t=2 \quad(\mathbf{D}) t=7 / 2 \quad(\mathbf{E}) t=4$$

Marginal Revenue Suppose the weekly revenue in dollars from selling x custom- made office desks is \(r(x)=2000\left(1-\frac{1}{x+1}\right)\) (a) Draw the graph of \(r .\) What values of \(x\) make sense in this problem situation? (b) Find the marginal revenue when \(x\) desks are sold. (c) Use the function \(r^{\prime}(x)\) to estimate the increase in revenue that will result from increasing sales from 5 desks a week to 6 desks a week. (d) Writing to Learn Find the limit of \(r^{\prime}(x)\) as \(x \rightarrow \infty\) How would you interpret this number?

Marginal Cost Suppose that the dollar cost of producing \(x\) washing machines is \(c(x)=2000+100 x-0.1 x^{2} .\) (a) Find the average cost of producing 100 washing machines. (b) Find the marginal cost when 100 machines are produced. (c) Show that the marginal cost when 100 washing machines are produced is approximately the cost of producing one more washing machine after the first 100 have been made, by calculating the latter cost directly.

Constant Acceleration Suppose the velocity of a falling body is \(v=k \sqrt{s} \mathrm{m} / \mathrm{sec}(k\) a constant) at the instant the body has fallen \(s\) meters from its starting point. Show that the body's acceleration is constant.
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