/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 The position of the front bumper... [FREE SOLUTION] | 91影视

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The position of the front bumper of a test car under microprocessor control is given by \(x(t) =\) 2.17 m \(+\) (4.80 m/s\(^2)t^2\) \(-\) (0.100 m/s\(^6)t^6\). (a) Find its position and acceleration at the instants when the car has zero velocity. (b) Draw \(x-t, v_x-t\), and \(a_x-t\) graphs for the motion of the bumper between \(t =\) 0 and \(t =\) 2.00 s.

Short Answer

Expert verified
The position at zero velocity instants is 2.17 m and 14.97 m; acceleration is 9.60 m/s虏 and -38.4 m/s虏, respectively.

Step by step solution

01

Find Velocity Function

To find when the car has zero velocity, calculate the derivative of the position function \(x(t)\) to get the velocity function \(v(t)\).The position function given is:\[x(t) = 2.17 + 4.80t^2 - 0.100t^6\]The velocity function is the derivative \(v(t) = \frac{d}{dt}(x(t))\):\[ v(t) = \frac{d}{dt}(2.17 + 4.80t^2 - 0.100t^6) = 9.60t - 0.600t^5\]
02

Set Velocity to Zero

Set the derived velocity function to zero and solve for \(t\):\[0 = 9.60t - 0.600t^5\]Factor out a \(t\):\[t(9.60 - 0.600t^4) = 0\]This gives two sets of solutions. Solving \(t = 0\) is obvious, while solving \(9.60 - 0.600t^4 = 0\):\[9.60 = 0.600t^4\]\[t^4 = 16\]\[t = \pm 2\]Since time cannot be negative, the times when velocity is zero are \(t = 0\) and \(t = 2\).
03

Calculate Position at Zero Velocity

Substitute the times \(t = 0\) and \(t = 2\) into the position function \(x(t)\) to find positions.For \(t = 0\):\[x(0) = 2.17 + 4.80(0)^2 - 0.100(0)^6 = 2.17\text{ m}\]For \(t = 2\):\[x(2) = 2.17 + 4.80(2)^2 - 0.100(2)^6 = 2.17 + 19.2 - 6.4 = 14.97\text{ m}\]
04

Find Acceleration Function

Acceleration is the derivative of the velocity function \(v(t)\).Taking the derivative:\[a(t) = \frac{d}{dt}(v(t)) = \frac{d}{dt}(9.60t - 0.600t^5) = 9.60 - 3.00t^4\]
05

Calculate Acceleration at Zero Velocity

Substitute \(t = 0\) and \(t = 2\) into the acceleration function \(a(t)\).For \(t = 0\):\[a(0) = 9.60 - 3.00(0)^4 = 9.60\text{ m/s}^2\]For \(t = 2\):\[a(2) = 9.60 - 3.00(2)^4 = 9.60 - 48 = -38.4\text{ m/s}^2\]
06

Graph the Functions

For graphing:- **Position vs. Time (\(x-t\))**: Plot the quadratic term dominating at early times, and the negative cubic term for later.- **Velocity vs. Time (\(v_x-t\))**: Shows a cubic characteristic with roots at \(t=0\) and \(t=2\).- **Acceleration vs. Time (\(a_x-t\))**: Remains positive early on, transitions negative after \(t=2\), showing a quartic decrease.Graphs provide a visual representation indicating changes at specific time-critical points.

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Key Concepts

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

Position Function
The position function describes where an object is located in space as a function of time. In kinematics, understanding the position of an object is pivotal, as it establishes the basis for analyzing motion.

In our example, the position function is given as:
  • \(x(t) = 2.17 + 4.80t^2 - 0.100t^6\)
This function tells us how the car's position changes over time. It consists of multiple terms: a constant term (2.17 m), a quadratic term (4.80t虏), and a sextic term (-0.100t鈦). Each component contributes differently:
  • The constant term sets an initial position.
  • The quadratic term suggests an accelerating motion, increasing the position over time.
  • The sextic term represents a more complex, higher-order influence that dominates at higher times, eventually decreasing the position as it involves a negative coefficient.
The position is calculated at specific time points, here when the car's velocity is zero, showing the spatial context of these moments.
Velocity Function
The velocity function provides insight into how the speed of an object changes over time. It is the derivative of the position function, representing the rate of change of position.

Deriving the position function, we have:
  • \(v(t) = \frac{d}{dt}(2.17 + 4.80t^2 - 0.100t^6) = 9.60t - 0.600t^5\)
Here:
  • The linear term (9.60t) suggests an increasing velocity over time, which implies acceleration.
  • The quintic term (-0.600t鈦) shows decreasing velocity at later times, revealing resistance or brakes in the motion.
The concept of zero velocity is crucial, as it indicates moments when the car stops momentarily before a possible change in direction. Solving \(v(t) = 0\) for time gives critical instants, which were found to be at \(t = 0\) and \(t = 2\) seconds in our example.
Acceleration Function
Acceleration is the measure of how quickly velocity changes. The acceleration function is derived from the velocity function, showing the rate of change of velocity.

Taking the derivative of the velocity function gives:
  • \(a(t) = \frac{d}{dt}(9.60t - 0.600t^5) = 9.60 - 3.00t^4\)
This function indicates both positive and negative phases of acceleration:
  • The constant term suggests initial positive acceleration.
  • The quartic term (-3.00t鈦) would dominate over time, resulting in negative acceleration, also known as deceleration.
At moments when velocity is zero, acceleration dictates the subsequent motion tendency. For instance, a positive acceleration at zero velocity may mean the car will start moving forward again, while negative acceleration can suggest reversal or slowing.
Graphical Analysis
Graphical analysis helps visualize and comprehend motion dynamics through plots of position, velocity, and acceleration against time. Such graphs provide a clear view of how motion parameters evolve.

- **Position vs. Time (\(x-t\))**: - This graph shows the trajectory of the object's position over time. The quadratic nature dominates initially, producing a parabolic path, and the negative sextic impact can be seen curving the graph downwards at larger times.- **Velocity vs. Time (\(v_x-t\))**: - Illustrates change in speed. Starting from zero, it hints at an accelerating phase followed by deceleration as shown by cubic damping. The graph crosses the horizontal axis at \(t = 0\) and \(t = 2\), marking periods of rest.- **Acceleration vs. Time (\(a_x-t\))**: - Depicts how acceleration changes. It begins positive, suggesting initial acceleration, but transitions to negative values, highlighting deceleration as time progresses. This graph helps identify when the car experiences braking forces.Each graph complements the analytical solutions, allowing for cross-verification of numerical results. They outline the entire scope of the car's kinematic behaviour between the selected time intervals.

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