Problem 39
Question
Velocity of a Particle The displacement \(s\) (in meters) of a particle moving in a straight line is given by the equation of motion \(s=4 t^{3}+6 t+2,\) where \(t\) is measured in seconds. Find the instantancous velocity of the particle \(s\) at times \(t=a, t=1, t=2, t=3\)
Step-by-Step Solution
Verified Answer
The instantaneous velocities are: \(v(a) = 12a^2 + 6\), \(v(1) = 18\), \(v(2) = 54\), \(v(3) = 114\).
1Step 1: Differentiate the displacement function
To find the instantaneous velocity, we need to differentiate the displacement function with respect to time. The displacement function is given by \(s(t) = 4t^3 + 6t + 2\). Applying the power rule of differentiation, we arrive at the velocity function: \(v(t) = \frac{ds}{dt} = 12t^2 + 6\).
2Step 2: Substitute values to find velocity at different times
Now that we have the velocity function \(v(t) = 12t^2 + 6\), we can find the instantaneous velocities at different time points by substituting these values into the velocity equation: - For \(t=a\), the velocity is \(v(a) = 12a^2 + 6\).- For \(t=1\), substitute \(t\) with 1: \(v(1) = 12(1)^2 + 6 = 12 + 6 = 18\).- For \(t=2\), substitute \(t\) with 2: \(v(2) = 12(2)^2 + 6 = 48 + 6 = 54\).- For \(t=3\), substitute \(t\) with 3: \(v(3) = 12(3)^2 + 6 = 108 + 6 = 114\).
Key Concepts
Displacement FunctionDifferentiationPower Rule
Displacement Function
The displacement function is essential in understanding how the position of a particle changes over time as it moves in a straight line. In the context of our exercise, the displacement function is represented as \( s(t) = 4t^3 + 6t + 2 \). This equation helps us determine the exact position of a particle at any given moment \( t \).
The formula combines different terms: a cubic term \( 4t^3 \), a linear term \( 6t \), and a constant term \( 2 \). Each of these has a unique role:
The formula combines different terms: a cubic term \( 4t^3 \), a linear term \( 6t \), and a constant term \( 2 \). Each of these has a unique role:
- The cubic term \( 4t^3 \) influences how quickly the displacement changes as time progresses.
- The linear term \( 6t \) alters the displacement based on time without involving any powers greater than one.
- The constant term \( 2 \) dictates the starting position of the particle.
Differentiation
Differentiation is the mathematical process used to find the rate of change of a function, which is crucial when trying to determine velocities from displacement functions. In this exercise, to find the instantaneous velocity, we differentiate the displacement function \( s(t) = 4t^3 + 6t + 2 \) with respect to time \( t \).
This process involves calculating the derivative, denoted by \( \frac{ds}{dt} \), which gives us the velocity function. Differentiating each term separately, we transform the cubic equation into a quadratic one, resulting in the velocity function \( v(t) = 12t^2 + 6 \). With this new function, you can easily substitute any given time \( t \) into it to determine the instantaneous velocity at that moment.
Differentiation turns a static description of a particle's trajectory into a dynamic expression of its motion.
This process involves calculating the derivative, denoted by \( \frac{ds}{dt} \), which gives us the velocity function. Differentiating each term separately, we transform the cubic equation into a quadratic one, resulting in the velocity function \( v(t) = 12t^2 + 6 \). With this new function, you can easily substitute any given time \( t \) into it to determine the instantaneous velocity at that moment.
Differentiation turns a static description of a particle's trajectory into a dynamic expression of its motion.
Power Rule
The power rule is one of the simplest but most useful rules in calculus, especially in differentiating polynomial functions like our displacement function. The power rule states that to differentiate a term \( at^n \), you multiply the term by its power and then decrease the power by one.
For the term \( 4t^3 \), using the power rule, you get \( 3 \times 4t^{3-1} = 12t^2 \), and for \( 6t \), you end up with \( 1 \times 6t^{1-1} = 6 \). Constants, like \( 2 \), disappear during differentiation because they do not change as time progresses.
This rule is not only efficient but also straightforward, enabling us to quickly differentiate complex polynomial expressions to find their derivative. Applying the power rule is fundamental to translating physical phenomena, like displacement of a particle, into mathematically manageable calculations like velocity.
For the term \( 4t^3 \), using the power rule, you get \( 3 \times 4t^{3-1} = 12t^2 \), and for \( 6t \), you end up with \( 1 \times 6t^{1-1} = 6 \). Constants, like \( 2 \), disappear during differentiation because they do not change as time progresses.
This rule is not only efficient but also straightforward, enabling us to quickly differentiate complex polynomial expressions to find their derivative. Applying the power rule is fundamental to translating physical phenomena, like displacement of a particle, into mathematically manageable calculations like velocity.
Other exercises in this chapter
Problem 38
Find the limit, if it exists. If the limit does not exist, explain why. $$\lim _{x \rightarrow-4^{-}} \frac{|x+4|}{x+4}$$
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Find the limit, if it exists. If the limit does not exist, explain why. $$\lim _{x \rightarrow 2} \frac{|x-2|}{x-2}$$
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(a) A tank contains \(5000 \mathrm{L}\) of pure water. Brine that contains \(30 \mathrm{g}\) of salt per liter of water is pumped into the tank at a rate of \(2
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