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Chapter 2: Problem 3

The velocity of a particle which moves along the \(s\) -axis is given by \(v=2-4 t+5 t^{3 / 2},\) where \(t\) is in seconds and \(v\) is in meters per second. Evaluate the position \(s,\) velocity \(v,\) and acceleration \(a\) when \(t=3\) s. The particle is at the position \(s_{0}=3 \mathrm{m}\) when \(t=0\)

Short Answer

Expert verified
Position: 43.39 m, Velocity: 15.98 m/s, Acceleration: 6.99 m/s虏 at t=3s.

Step by step solution

01

Calculate Velocity at t=3s

Substitute \(t=3\) into the velocity equation \(v = 2 - 4t + 5t^{3/2}\):\[ v = 2 - 4(3) + 5(3)^{3/2} \]This simplifies to:\[ v = 2 - 12 + 5(3\sqrt{3}) \]Calculate further:\[ v = -10 + 15\sqrt{3} \]Thus, the velocity is approximately \(v = 15.98 \text{ m/s}\)
02

Calculate Acceleration at t=3s

Find the derivative of the velocity function to get the acceleration function:\[ a(t) = \frac{d}{dt}(2 - 4t + 5t^{3/2}) \]Apply the power rule:\[ a(t) = 0 - 4 + \frac{15}{2}t^{1/2} \]Now substitute \(t = 3\):\[ a(3) = -4 + \frac{15}{2}(3^{1/2}) \]Calculate further:\[ a(3) = -4 + \frac{15\sqrt{3}}{2} \]Thus, the acceleration is approximately \(a = 6.99 \text{ m/s}^2\).
03

Find the Position at t=3s

Integrate the velocity function to find the position function \(s(t)\):\[ s(t) = \int (2 - 4t + 5t^{3/2}) dt \]Calculate the indefinite integral:\[ s(t) = 2t - 2t^2 + \frac{10}{5} \cdot \frac{2}{5} t^{5/2} + C = 2t - 2t^2 + 4t^{5/2} + C \]Use initial condition \(s(0) = 3\) to find \(C\):\[ 3 = 2(0) - 2(0)^2 + 4(0)^{5/2} + C \Rightarrow C = 3 \]For \(t = 3\), substitute into the position function:\[ s(3) = 2(3) - 2(3)^2 + 4(3)^{5/2} + 3 \]Calculate further:\[ s(3) = 6 - 18 + 12(3\sqrt{3}) + 3 \approx 43.392 \]Thus, the position is approximately \(s = 43.39 \text{ m}\) at \(t = 3\) s.

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Key Concepts

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

Velocity Calculation
Understanding how to calculate velocity is crucial in particle dynamics. Velocity is essentially the speed of a particle in a specific direction. In the given problem, the velocity equation provided is
  • \( v = 2 - 4t + 5t^{3/2} \),
where \( v \) is the velocity and \( t \) is time in seconds. To determine the velocity at a specific time, substitute the value of \( t \) into the velocity equation. For example, when \( t = 3 \) seconds, this substitution looks like:
  • \( v = 2 - 4 \times 3 + 5 \times (3)^{3/2} \)
This simplifies to:
  • \( v = -10 + 15\sqrt{3} \),
yielding a velocity of approximately 15.98 m/s. Calculating velocity involves replacing \( t \) in the equation and simplifying, which gives insight into the speed and direction the particle is traveling at any point in time.
Acceleration Determination
Acceleration reveals how a particle's velocity changes over time, crucial for predicting movement patterns. To determine acceleration, the derivative of the velocity function is needed. For our scenario:
  • Start with the velocity function: \( v = 2 - 4t + 5t^{3/2} \).
  • Find the derivative to determine acceleration: \( a(t) = \frac{d}{dt}(2 - 4t + 5t^{3/2}) \).
Applying the power rule gives us:
  • \( a(t) = 0 - 4 + \frac{15}{2}t^{1/2} \).
Then, by substituting \( t = 3 \) seconds, we calculate:
  • \( a(3) = -4 + \frac{15\sqrt{3}}{2} \),
resulting in an acceleration of approximately 6.99 m/s虏. This calculation provides insight into how quickly the particle's velocity is changing, which is vital for understanding the dynamics of its motion.
Position Analysis
Position analysis involves understanding where the particle is at a given time. To find the position function \( s(t) \), you integrate the velocity function:
  • Start with \( s(t) = \int (2 - 4t + 5t^{3/2}) dt \).
The integral results in:
  • \( s(t) = 2t - 2t^2 + 4t^{5/2} + C \).
Using the initial condition \( s(0) = 3 \), find constant \( C \):
  • With \( t = 0 \), you solve \( 3 = 2(0) - 2(0)^2 + 4(0)^{5/2} + C \), concluding \( C = 3 \).
Finally, substitute \( t = 3 \) into the position function:
  • Calculate: \( s(3) = 2(3) - 2(3)^2 + 4(3)^{5/2} + 3 \).
  • Simplifying gives: \( s(3) = 6 - 18 + 12(3\sqrt{3}) + 3 \),
resulting in a position of approximately 43.39 m. Position analysis helps in understanding the exact location of the particle at any time \( t \).

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Most popular questions from this chapter

The radial position of a fluid particle \(P\) in a certain centrifugal pump with radial vanes is approximated by \(r=r_{0} \cosh K t,\) where \(t\) is time and \(K=\dot{\theta}\) is the constant angular rate at which the impeller turns. Determine the expression for the magnitude of the total acceleration of the particle just prior to leaving the vane in terms of \(r_{0}, R,\) and \(K\)

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The acceleration of a particle is given by \(a=2 t-10\) where \(a\) is in meters per second squared and \(t\) is in seconds. Determine the velocity and displacement as functions of time. The initial displacement at \(t=0\) is \(s_{0}=-4 \mathrm{m},\) and the initial velocity is \(v_{0}=3 \mathrm{m} / \mathrm{s}\)
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