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Chapter 2: Problem 4

The displacement of a particle which moves along the \(s\) -axis is given by \(s=(-2+3 t) e^{-0.5 t},\) where \(s\) is in meters and \(t\) is in seconds. Plot the displacement, velocity, and acceleration versus time for the first 20 seconds of motion. Determine the time at which the acceleration is zero.

Short Answer

Expert verified
Acceleration is zero at approximately 4.67 seconds.

Step by step solution

01

Plot the Displacement Function

To plot the displacement, use the function \(s(t) = (-2 + 3t) e^{-0.5t}\). Substitute values of \(t\) from 0 to 20 in increments (e.g. 0.5 seconds) to calculate \(s\) and plot \(s\) against \(t\). The plot will show how displacement varies with time.
02

Differentiate to Find Velocity

The velocity \(v(t)\) is the first derivative of the displacement function with respect to time \(t\). Find \(v(t)\) by differentiating \(s(t)\):\[v(t) = \frac{d}{dt} [(-2 + 3t) e^{-0.5t}] = (3 - 0.5(-2 + 3t)) e^{-0.5t}\]Simplify to get \(v(t) = (3 - 1.5t + 1) e^{-0.5t} = (4 - 1.5t) e^{-0.5t}\). Plot \(v(t)\) against \(t\) between 0 and 20 seconds.
03

Differentiate to Find Acceleration

Acceleration \(a(t)\) is the second derivative of the displacement function with respect to time \(t\). Differentiate \(v(t) = (4 - 1.5t) e^{-0.5t}\) to find:\[a(t) = \frac{d}{dt}[(4 - 1.5t) e^{-0.5t}] = (-0.5(4 - 1.5t) - 1.5) e^{-0.5t}\]Simplify to get: \[a(t) = (-2 + 0.75t - 1.5) e^{-0.5t} = (-3.5 + 0.75t) e^{-0.5t}\].Plot \(a(t)\) against time \(t\) from 0 to 20 seconds.
04

Solve for Zero Acceleration

To find when acceleration \(a(t)\) is zero, solve the equation:\[-3.5 + 0.75t = 0\]\[0.75t = 3.5\]\[t = \frac{3.5}{0.75} = \frac{35}{7.5} = 4.6667\].Thus, the acceleration is zero at approximately \(t = 4.67\) seconds.

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Key Concepts

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

Differentiation
Differentiation is a powerful mathematical process used to calculate the rate at which quantities change. In the context of motion, it is applied to the displacement function to find velocity and acceleration.
The given displacement function is \(s(t) = (-2 + 3t) e^{-0.5t}\). To determine how fast the displacement changes, we differentiate this function with respect to time \(t\).
This is done through the product rule for differentiation and results in the velocity function. This tells us how quickly and in what direction the particle is moving at any given point in time.
Velocity
Velocity is the first derivative of the displacement function, and it tells us how quickly a particle moves and in which direction. By differentiating the displacement function \(s(t)\), we obtained the velocity function:
\[v(t) = (4 - 1.5t) e^{-0.5t}\]
Here, \(v(t)\) indicates how the speed of the particle changes over time. For instance, if you plug in values of \(t\), you鈥檒l see when the velocity is positive or negative, representing forward and backward motion, respectively.
Plotting velocity against time helps visualize these changes over a period like 0 to 20 seconds.
Acceleration
Acceleration is the rate at which the velocity changes with time. It is derived by differentiating the velocity function. For our exercise, applying differentiation on \(v(t) = (4 - 1.5t) e^{-0.5t}\) results in:
\[a(t) = (-3.5 + 0.75t) e^{-0.5t}\]
This new function tells us how the particle鈥檚 velocity is accelerating or decelerating over time. In other words, acceleration helps understand if the particle is speeding up or slowing down as it moves along the \(s\)-axis.
Zero Acceleration
Zero acceleration indicates a moment when there is no change in velocity. In other words, the particle is momentarily moving at a constant speed. To find this point, set the acceleration function \(a(t) = (-3.5 + 0.75t) e^{-0.5t}\) to zero and solve for \(t\).
The solution is \(t \approx 4.67\) seconds, meaning at this time, the particle's velocity is unchanging. This critical point can signify a switch from speeding up to slowing down or a constant velocity period.
Plotting Functions
Plotting functions is an essential part of visualizing and understanding physical motion. By graphing displacement, velocity, and acceleration against time, we gain insights into how a particle moves.
For our problem, we chart these functions from \(t = 0\) to \(t = 20\) seconds. Observing these graphs can show peaks, slopes, and periods of rest, which tell stories of how the particle behaves.
  • Displacement Plot: Shows the overall path or position of the particle over time.
  • Velocity Plot: Indicates how fast and in which direction the particle moves.
  • Acceleration Plot: Highlights the changes in the particle's velocity, depicting moments of speeding up or slowing down.
Understanding these plots aids in interpreting complex motion easily and engagingly.

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Most popular questions from this chapter

The car \(A\) is ascending a parking-garage ramp in the form of a cylindrical helix of 24 -ft radius rising 10 ft for each half turn. At the position shown the car has a speed of \(15 \mathrm{mi} / \mathrm{hr}\), which is decreasing at the rate of \(2 \mathrm{mi} / \mathrm{hr}\) per second. Determine the \(r, \theta\) and \(z\) -components of the acceleration of the car.

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 \(t=0\)

A test car starts from rest on a horizontal circular track of \(80-\mathrm{m}\) radius and increases its speed at a uniform rate to reach \(100 \mathrm{km} / \mathrm{h}\) in 10 seconds. Determine the magnitude \(a\) of the total acceleration of the car 8 seconds after the start.

The speed of a car increases uniformly with time from \(50 \mathrm{km} / \mathrm{h}\) at \(A\) to \(100 \mathrm{km} / \mathrm{h}\) at \(B\) during \(10 \mathrm{sec}\) onds. The radius of curvature of the hump at \(A\) is \(40 \mathrm{m} .\) If the magnitude of the total acceleration of the mass center of the car is the same at \(B\) as at \(A,\) compute the radius of curvature \(\rho_{B}\) of the dip in the road at \(B\). The mass center of the car is \(0.6 \mathrm{m}\) from the road.

A particle in an experimental apparatus has a velocity given by \(v=k \sqrt{s},\) where \(v\) is in millimeters per second, the position \(s\) is millimeters, and the constant \(k=0.2 \mathrm{mm}^{1 / 2} \mathrm{s}^{-1}\). If the particle has a velocity \(v_{0}=3 \mathrm{mm} / \mathrm{s}\) at \(t=0,\) determine the particle position, velocity, and acceleration as functions of time, and compute the time, position, and acceleration of the particle when the velocity reaches \(15 \mathrm{mm} / \mathrm{s}\)
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