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Chapter 23: Problem 38

Find the acceleration of an object for which the displacement \(s\) (in \(\mathrm{m}\) ) is given as a function of the time \(t\) (in s) for the given value of \(t\). $$s=3(1+2 t)^{4}, t=0.500 \mathrm{s}$$

Short Answer

Expert verified
The acceleration at \( t = 0.500 \) s is \( 576 \, \text{m/s}^2 \).

Step by step solution

01

Find the Velocity Function

To find the acceleration, we first need to determine the velocity function, which is the derivative of the displacement function with respect to time. The displacement function is given by \( s = 3(1+2t)^4 \). Applying the chain rule, we differentiate to get the velocity function: \[ v(t) = \frac{ds}{dt} = 3 \cdot 4(1+2t)^3 \cdot 2 = 24(1+2t)^3 \]
02

Find the Acceleration Function

The acceleration is the derivative of the velocity function with respect to time. We have found the velocity function to be \( v(t) = 24(1+2t)^3 \). Differentiating this with respect to time, again using the chain rule, gives:\[ a(t) = \frac{dv}{dt} = 24 \cdot 3(1+2t)^2 \cdot 2 = 144(1+2t)^2 \]
03

Evaluate the Acceleration at the Given Time

Finally, substitute the given time value \( t = 0.500 \) s into the acceleration function to find the specific acceleration at this time:\[ a(0.500) = 144(1+2 \cdot 0.500)^2 = 144(2)^2 = 144 \cdot 4 = 576 \, \text{m/s}^2 \]

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Key Concepts

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

Chain Rule
The chain rule is a fundamental concept in calculus. It helps us differentiate composite functions. Imagine we have a function within another function, like an onion with layers. The chain rule tells us how to differentiate a function like this.
To apply it, follow these steps:
  • Differentiate the outer function.
  • Multiply it by the derivative of the inner function.
In our example, the displacement function is given by \(s = 3(1+2t)^4\). Here, the inner function is \(1 + 2t\), and the outer function is raising it to the fourth power, \((1+2t)^4\).
We apply the chain rule by differentiating the outer function first and then multiplying by the derivative of the inner function, creating a sequence of steps for breaking down complex derivatives into manageable parts.
Derivative
The concept of a derivative is central to calculus. It represents how a quantity changes with respect to change in another quantity. Think of it like the slope of a curve at a given point.
When you find the derivative, you're identifying the rate of change. For example, the derivative of displacement with respect to time is the velocity, while the derivative of velocity with respect to time is acceleration.
  • Start with your function, like displacement \(s = 3(1+2t)^4\).
  • Apply rules like the power rule or chain rule to find its derivative.
  • For \(s = 3(1+2t)^4\), this means first finding \(v(t) = \frac{ds}{dt}\).
  • Then, derive again to find acceleration: \(a(t) = \frac{dv}{dt}\).
Through derivatives, we unlock the door to understanding how motion and other physical phenomena evolve with time.
Acceleration
Acceleration is the rate at which velocity changes with time. It's a measure of how quickly something speeds up or slows down.
When you calculate acceleration, you're finding the derivative of the velocity function with respect to time.
  • From our example, we start with the velocity function \(v(t) = 24(1+2t)^3\).
  • To find the acceleration, apply the chain rule again to get \(a(t) = 144(1+2t)^2\).
  • Evaluate this at the specific point \(t = 0.500\) s to find \(a(0.500) = 576 \, \text{m/s}^2\).
Understanding acceleration helps in predicting how the speed of an object changes with time, which is vital in physics and engineering.
Velocity Function
The velocity function measures the speed of an object and the direction it's moving in, at any given instant of time. It's effectively the first derivative of the position or displacement function with respect to time.
In our problem, we start with the displacement function \(s = 3(1+2t)^4\) and find its derivative to get the velocity function.
  • The process involves applying the chain rule, resulting in \(v(t) = 24(1+2t)^3\).
  • This tells us how fast the position is changing as a function of time, revealing that velocity is a dynamic, changing quantity.
The velocity function is crucial for assessing how objects move, allowing us to predict motion patterns and calculate other important aspects like acceleration.

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Most popular questions from this chapter

Solve the given problems involving limits. The area \(A\) (in \(\mathrm{mm}^{2}\) ) of the pupil of a certain person's eye is given by \(A=\frac{36+24 b^{3}}{1+4 b^{3}},\) where \(b\) is the brightness (in lumens) of the light source. Between what values does \(A\) vary?

Solve the given problems by finding the appropriate derivatives.The deflection \(y\) (in \(\mathrm{m}\) ) of a \(5.00-\mathrm{m}\) beam as a function of the distance \(x\) (in \(\mathrm{m}\) ) from one end is \(y=0.0001\left(x^{5}-25 x^{2}\right) .\) Find the value of \(d^{2} y / d x^{2}\) (the rate of change at which the slope of the beam changes) where \(x=3.00 \mathrm{m}\).

\(\lim _{x \rightarrow a^{-}} f(x)\) means to find the limit as x approaches a from the left only, and \(\lim _{x \rightarrow a^{+}} f(x)\) means to find the limit as \(x\) approaches a from the right only. These are called one-sided limits. Solve the following problems. In Einstein's theory of relativity, the length \(L\) of an object moving at a velocity \(v\) is \(L=L_{0} \sqrt{1-\frac{v^{2}}{c^{2}}},\) where \(c\) is the speed of light and \(L_{0}\) is the length of the object at rest. Find lim \(L\) and explain why a limit from the left is used.

Solve the given problems by using implicit differentiation.Find the slope of a line tangent to the curve of the implicit function \(x y+y^{2}+2=0\) at the point \((-3,1) .\) Use the derivative evaluation feature of a calculator to check your result.

\(\lim _{x \rightarrow a^{-}} f(x)\) means to find the limit as x approaches a from the left only, and \(\lim _{x \rightarrow a^{+}} f(x)\) means to find the limit as \(x\) approaches a from the right only. These are called one-sided limits. Solve the following problems. Is there a difference between \(\lim _{x \rightarrow 2^{-}} \frac{1}{x-2}\) and \(\lim _{x \rightarrow 2^{+}} \frac{1}{x-2} ?\)
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