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Chapter 1: Problem 48

Given \(s(t)=t^{2}-\frac{1}{2} t+3\) where \(s(t)\) is in meters and \(t\) is in seconds, find each of the following. a) \(v(t)\) b) \(a(t)\) c) The velocity and acceleration when \(t=1 \sec\)

Short Answer

Expert verified
a) Velocity function: \(v(t) = 2t - \frac{1}{2}\). b) Acceleration function: \(a(t) = 2\). c) At \(t=1\): Velocity = 1.5 m/s and Acceleration = 2 m/s虏.

Step by step solution

01

Find the Velocity Function

The velocity function, \(v(t)\), is the first derivative of the position function \(s(t)\) with respect to time \(t\). To find \(v(t)\), differentiate \(s(t) = t^2 - \frac{1}{2}t + 3\) with respect to \(t\). The derivative is:\[v(t) = \frac{d}{dt}(t^2) - \frac{d}{dt}\left(\frac{1}{2}t\right) + \frac{d}{dt}(3)\]Calculating each part:- \(\frac{d}{dt}(t^2) = 2t\)- \(\frac{d}{dt}(\frac{1}{2}t) = \frac{1}{2}\)- \(\frac{d}{dt}(3) = 0\)Thus, \(v(t) = 2t - \frac{1}{2}\).
02

Find the Acceleration Function

The acceleration function, \(a(t)\), is the first derivative of the velocity function \(v(t)\) or the second derivative of the position function \(s(t)\). Differentiate \(v(t) = 2t - \frac{1}{2}\) with respect to \(t\):\[a(t) = \frac{d}{dt}(2t) - \frac{d}{dt}\left(\frac{1}{2}\right)\]Calculating each part:- \(\frac{d}{dt}(2t) = 2\)- \(\frac{d}{dt}(\frac{1}{2}) = 0\)Thus, \(a(t) = 2\).
03

Calculate Velocity at \(t = 1\) Second

Using the velocity function \(v(t) = 2t - \frac{1}{2}\), find \(v(1)\):\[v(1) = 2(1) - \frac{1}{2} = 2 - 0.5 = 1.5\text{ m/s}\]
04

Calculate Acceleration at \(t = 1\) Second

Since the acceleration function \(a(t)\) is constant and equal to 2, the acceleration at any time \(t\) is the same. Therefore:\[a(1) = 2\text{ m/s}^2\]

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Key Concepts

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

Derivatives
In calculus, derivatives are a fundamental concept that represent the rate at which a quantity changes. When we have a position function like \(s(t) = t^2 - \frac{1}{2}t + 3\), which describes how the position of an object changes over time, the derivative with respect to time \(t\) gives us the velocity. The process of differentiation shows us how small changes in time result in changes in position.
  • To find a derivative, we apply differentiation rules to each term of the function separately.
  • For example, the derivative of \(t^2\) is \(2t\), reflecting that its rate of change doubles with every unit increase in time.
  • Constants, like \(3\), have a derivative of zero because they do not change.
Thus, to find the velocity function, differentiate the position function \(s(t)\) with respect to time. The resulting function \(v(t) = 2t - \frac{1}{2}\) tells us how quickly the object's position is changing at any given time t.
Velocity
Velocity is all about how fast an object is moving and in which direction. It's a vector quantity, meaning it has both magnitude and direction, unlike speed, which only has magnitude. The velocity function, derived from the position function, informs us about these changes over time.
  • For our exercise, the velocity function is \(v(t) = 2t - \frac{1}{2}\).
  • This tells us the velocity of the object changes linearly as time changes.
  • At \(t = 1\) second, substituting into the velocity equation gives \(v(1) = 1.5\) meters per second.
This means at that specific point in time, the object is moving at 1.5 meters per second in whatever direction is specified.
Acceleration
Acceleration is the rate at which velocity changes over time. Just like velocity is the derivative of position, acceleration is the derivative of velocity. It tells us if an object is speeding up, slowing down, or changing direction.
  • In our case, the acceleration is derived from the velocity function.
  • The acceleration function \(a(t) = 2\) is constant, which means the change in velocity is the same at every point in time, indicating uniform acceleration.
  • At \(t = 1\) second, the acceleration remains \(2\) meters per second squared, showing a constant increase in speed.
A constant acceleration simplifies many calculations in physics and engineering because it implies the object's movement is predictable and regular over time.

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Most popular questions from this chapter

Find an equation of the tangent line to the graph of \(f(x)=\frac{1}{x^{2}}\) a) at (1,1)\(;\) b) at \(\left(3, \frac{1}{9}\right)\) c) at \(\left(-2, \frac{1}{4}\right)\).

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ f(x)=\frac{1}{3} x^{3}+\frac{1}{2} x^{2}-2 $$

It has been shown that the home range, in hectares, of a carnivorous mammal weighing \(w\) grams can be approximated by $$ H(w)=0.11 w^{1.36} $$ (Source: Based on information in Emlen, J. M., Ecology: An Evolutionary Approach, p. \(216,\) Reading, MA: Addison-Wesley, 1973 ; and Harestad, A. S., and Bunnel, F. L., "Home Range and Body Weight-A Reevaluation," Ecology, Vol. 60, No. 2, pp. 405-418.) a) Find the average rate at which a carnivorous mammal's home range increases as the animal's weight grows from \(500 \mathrm{~g}\) to \(700 \mathrm{~g}\). b) Find \(\frac{H(300)-H(200)}{300-200}\). What does this rate represent?

Graph the function \(f\) given by $$ f(x)=\left\\{\begin{array}{ll} -3, & \text { for } x=-2 \\ x^{2}, & \text { for } x \neq-2 \end{array}\right. $$ Use GRAPH and TRACE to find each of the following limits. When necessary, state that the limit does not exist. a) \(\lim _{x \rightarrow-2^{+}} f(x)\) b) \(\lim _{x \rightarrow-2^{-}} f(x)\) c) \(\lim _{x \rightarrow-2} f(x)\) d) \(\lim _{x \rightarrow 2^{+}} f(x)\) e) \(\lim _{x \rightarrow 2^{-}} f(x)\) f) Does \(\lim _{x \rightarrow-2} f(x)=f(-2)\) ? g) Does \(\lim _{x \rightarrow 2} f(x)=f(2)\) ?

For each function, find the points on the graph at which the tangent line has slope 1 . $$ y=-0.01 x^{2}+2 x $$
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