91Ó°ÊÓ

The position of a particle moving along the \(x\) axis depends on the time according to the equation \(x=c t^{2}-b t^{3}\), where \(x\) is in meters and \(t\) in seconds. What are the units of (a) constant \(c\) and (b) constant \(b\) ? Let their numerical values be \(3.0\) and \(2.0\), respectively. (c) At what time does the particle reach its maximum positive \(x\) position? From \(t=0.0 \mathrm{~s}\) to \(t=4.0 \mathrm{~s},(\mathrm{~d})\) what distance does the particle move and (e) what is its displacement? Find its velocity at times (f) \(1.0 \mathrm{~s},(\mathrm{~g}) 2.0 \mathrm{~s},(\mathrm{~h}) 3.0 \mathrm{~s}\), and (i) \(4.0 \mathrm{~s}\). Find its acceleration at times (j) \(1.0 \mathrm{~s},(\mathrm{k}) 2.0 \mathrm{~s},(1) 3.0 \mathrm{~s}\), and \((\mathrm{m}) 4.0 \mathrm{~s}\).

Step by step solution

01

Determine Units for Constant c

Since the position \(x\) is given in meters, the unit for \(x\) must also be meters. In the term \(ct^2\), \(t\) is time in seconds squared, so \(c\) must have units of \(\text{m/s}^2\) in order for \(ct^2\) to result in meters.

02

Determine Units for Constant b

In the term \(bt^3\), \(t\) is time in seconds cubed, so \(b\) must have units of \(\text{m/s}^3\) to ensure \(bt^3\) results in meters. Therefore, the units of \(b\) are \(\text{m/s}^3\).

03

Time at Maximum x

To find the time at which the particle reaches its maximum positive position, we need to find when the velocity is zero. The velocity \(v(t)\) is the derivative of the position function: \(v = \frac{d}{dt}(ct^2 - bt^3) = 2ct - 3bt^2\). Set \(v = 0\) to find the critical points: \(2ct = 3bt^2\) simplifies to \(2c = 3bt\), thus \(t = \frac{2c}{3b}\). Substitute \(c = 3.0\) and \(b = 2.0\), \(t = \frac{2\times3}{3\times2} = 1.0\) second.

04

Distance Moved from t=0 to t=4 s

The distance is the total path traveled, which can be different from displacement. Here, compute positions at critical points to check changes in direction. Evaluate \(x(t)\) at \(t = 0, 1, 4\), finding positions. We need the absolute difference: \(x(0) = 0\), \(x(1) = 3\), \(x(4) = -80\). Calculate total distance: \(|x(1) - x(0)| + |x(4) - x(1)| = 3 + 83 = 86\) meters.

05

Displacement from t=0 to t=4 s

Displacement is simply \(x(4) - x(0) = -80 - 0 = -80\) meters.

06

Velocity at Specific Times (1s, 2s, 3s, 4s)

Compute velocity \(v(t) = 2ct - 3bt^2\) and substitute given times. - At \(t=1:\) \(v = 2(3)\cdot 1 - 3(2)\cdot 1^2 = 6 - 6 = 0\) m/s.- At \(t=2:\) \(v = 2(3)\cdot 2 - 3(2)\cdot 4 = 12 - 24 = -12\) m/s.- At \(t=3:\) \(v = 2(3)\cdot 3 - 3(2)\cdot 9 = 18 - 54 = -36\) m/s.- At \(t=4:\) \(v = 2(3)\cdot 4 - 3(2)\cdot 16 = 24 - 96 = -72\) m/s.

07

Acceleration at Specific Times (1s, 2s, 3s, 4s)

Acceleration is the derivative of velocity: \(a(t) = \frac{d}{dt}(2ct - 3bt^2) = 2c - 6bt\). Substitute times:- At \(t=1:\) \(a = 6 - 12 = -6\) m/s².- At \(t=2:\) \(a = 6 - 24 = -18\) m/s².- At \(t=3:\) \(a = 6 - 36 = -30\) m/s².- At \(t=4:\) \(a = 6 - 48 = -42\) m/s².

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Key Concepts

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

Position Equation

Understanding the position equation is the foundation for analyzing particle motion. This equation describes the position of a particle along a given axis over time. In the form of \( x = c t^2 - b t^3 \), this quadratic and cubic expression allows us to predict where a particle will be at any given moment, as time progresses.

Each term in the equation has a distinct role:

• \(c t^2\): This term represents motion that has initially increasing speed, quadratically dependent on time.
• \(b t^3\): This term implies a deceleration in motion, because the value is subtracted, and it is cubic in relation to time.

Knowing how to interpret and manipulate these parts is critical in physics to predict and understand the motion of particles or objects. This fundamental position equation guides us in analyzing further velocity and acceleration.

Derivatives in Physics

Derivatives in physics serve to find the rate of change of quantities, like velocity and acceleration from position variables. By differentiating the position equation \( x = c t^2 - b t^3 \) with respect to time \( t \), we derive the velocity equation.

Velocity is given by:

• \( v(t) = \frac{d}{dt}(c t^2 - b t^3) = 2ct - 3bt^2 \)

This equation shows how velocity changes as time progresses, as derived from the position equation. It's the first derivative of position.

To find acceleration, we differentiate velocity:

• \( a(t) = \frac{d}{dt}(2ct - 3bt^2) = 2c - 6bt \)

Understanding these derivatives allows us to translate position, velocity, and acceleration into actionable physical understanding and prediction within motion scenarios.

Units Conversion

Units are essential in ensuring mathematical calculations have real-world meaning. In physics, matching units can confirm the correctness of equations or guide us in finding constants like \( c \) and \( b \) in a position equation. For the given function \( x = c t^2 - b t^3 \), the units of \( x \) dictate the required units for constants \( c \) and \( b \).

• For \( c \): Since \( x \) is in meters and \( t^2 \) in seconds squared \((s^2)\), \( c \) must be in \( \text{m/s}^2 \).
• For \( b \): Similarly, with \( t^3 \) in seconds cubed \((s^3)\), \( b \) should be in \( \text{m/s}^3 \).

By comprehending proper units conversion, we can maintain consistency in formulas and properly interpret quantitative results in physical terms, valid across different scenarios and conditions.

Velocity and Acceleration Calculations

To grasp how a particle moves, compute velocity and acceleration at specific times using derivatives. Given the position equation \( x = c t^2 - b t^3 \), velocity \( v(t) \) is the first derivative, and it captures speed and direction changes.

For different moments like \(t = 1s, 2s, 3s, 4s\):

• At \( t=1 \): \( v = 0 \) m/s, meaning no speed at that point.
• At \( t=2 \): \( v = -12 \) m/s, motion reverses or slows significantly.
• At \( t=3 \): \( v = -36 \) m/s, speeds up in the negative direction.
• At \( t=4 \): \( v = -72 \) m/s, further acceleration in reverse.

For acceleration \( a(t) \), derived from velocity:

• Each point demonstrates increasing negative acceleration, indicating continual decrease in velocity over time.
• Evaluation at \( t=1, 2, 3, 4 \) gives values of \( -6, -18, -30, -42 \ m/s^2 \) respectively.

Interpreting these calculated values helps us predict how fast and in what manner a particle accelerates, illustrating complex motion behavior effectively.