Problem 22 - Fundamentals of Physics

Question

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 valucs be 3.0 and \(2.0,\) respectivcly. (c) At what time does the particle reach its maximum positive \(x\) position? From \(t=0.0 \mathrm{~s}\) to \(t=4.0 \mathrm{~s},\) (d) what distance does the particle move and (e) what is its displacement? Find its velocity at times (f) \(1.0 \mathrm{~s},(\mathrm{~g}) 2.0 \mathrm{~s},(\mathrm{~h}) \overline{3.0 \mathrm{~s}, \text { and }}\) (i) \(4.0 \mathrm{~s}\). Find its acceleration at times (j) \(1.0 \mathrm{~s},(\mathrm{k}) 2.0 \mathrm{~s},(\mathrm{l}) 3.0 \mathrm{~s},\) and \((\mathrm{m}) 4.0 \mathrm{~s}\)

Step-by-Step Solution

Answer

(a) m/s², (b) m/s³; (c) t = 1 s; (d) 80 m; (e) -80 m; (f) 0 m/s, (g) -12 m/s, (h) -36 m/s, (i) -72 m/s; (j) -6 m/s², (k) -12 m/s², (l) -30 m/s², (m) -42 m/s².

Step 1: Determine the units of constants c and b

The equation for the position is given by \( x = ct^2 - bt^3 \). Since \( x \) is in meters, each term on the right side must also be in meters. For \( ct^2 \), the units of \( c \) must cancel \( t^2 \) to result in meters. Thus, \([c] = \text{meters/seconds}^2\). For \( bt^3 \), the units of \( b \) must cancel \( t^3 \) to result in meters, so \([b] = \text{meters/seconds}^3\).

Step 2: Determine the time when x is maximum

To find when the particle reaches its maximum position, take the derivative of \( x \) with respect to \( t \) to find the velocity, \( v = \frac{dx}{dt} = 2ct - 3bt^2 \). Set \( v = 0 \) and solve for \( t \). This gives \( 2ct = 3bt^2 \), and solving for \( t \), we get \( t = \frac{2c}{3b} \). Substituting \( c = 3.0 \) and \( b = 2.0 \), \( t = \frac{2(3)}{3(2)} = 1.0 \text{ s} \).

Step 3: Calculate the distance moved from t = 0.0 s to t = 4.0 s

The distance moved is found by integrating the absolute value of velocity from \( t = 0.0 \) s to \( t = 4.0 \) s or by checking particle direction change intervals. Calculate \( x \) at these times: \( x(0) = 0 \), \( x(4) = 3(4)^2 - 2(4)^3 = 48 - 128 = -80 \text{ meters} \). A change in sign in velocity indicates a change in direction, so additional calculations are required to find distinct positions and absolute values added.

Step 4: Calculate the displacement from t = 0.0 s to t = 4.0 s

The displacement is the straight line distance from the initial to the final position of the particle, regardless of the path taken. Thus, displacement is \( x(4) - x(0) = -80 - 0 = -80 \text{ meters} \).

Step 5: Calculate velocity at specified times

The expression for velocity is \( v = 2ct - 3bt^2 \). Substitute the values of \( c \) and \( b \) and times to find: (f) \( v(1) = 2(3)(1) - 3(2)(1)^2 = 6 - 6 = 0 \text{ m/s} \), (g) \( v(2) = 2(3)(2) - 3(2)(2)^2 = 12 - 24 = -12 \text{ m/s} \), (h) \( v(3) = 2(3)(3) - 3(2)(3)^2 = 18 - 54 = -36 \text{ m/s} \), and (i) \( v(4) = 2(3)(4) - 3(2)(4)^2 = 24 - 96 = -72 \text{ m/s} \).

Step 6: Calculate acceleration at specified times

The expression for acceleration is the derivative of velocity, \( a = \frac{dv}{dt} = 2c - 6bt \). Substitute the values of \( c \) and \( b \) and the specific times to find: (j) \( a(1) = 2(3) - 6(2)(1) = 6 - 12 = -6 \text{ m/s}^2 \), (k) \( a(2) = 6 - 6(2) = -12 \text{ m/s}^2 \), (l) \( a(3) = 6 - 6(6) = -30 \text{ m/s}^2 \), and (m) \( a(4) = 6 - 6(8) = -42 \text{ m/s}^2 \).

Key Concepts

Velocity

Velocity describes how fast an object is moving and in which direction. In this exercise, the particle's velocity is given by the derivative of its position equation. This means the velocity equation is found by differentiating the position equation, \( x = ct^2 - bt^3 \), with respect to time \( t \). The result is \( v(t) = \frac{dx}{dt} = 2ct - 3bt^2 \). Here, velocity depends on time and is affected by both constants \( c \) and \( b \).
- **At different times**, you can simply substitute \( t \) into the velocity equation to find the velocity at that particular time.
- For example, the velocity when \( t = 1 \) second is \( v(1) = 2c(1) - 3b(1)^2 \). This allows us to check if the particle is speeding up, slowing down, or changing direction based on whether \( v(t) \) is positive, negative, or zero. At \( t = 1 \text{ s} \) the velocity is \( 0 \text{ m/s} \), indicating a momentary stop or change in direction.

Acceleration

Acceleration is the rate of change of velocity with respect to time. It indicates how quickly the velocity of an object is changing. To find the acceleration from a velocity equation, differentiate the velocity equation with respect to time.
- Given the velocity equation \( v = 2ct - 3bt^2 \), the acceleration \( a \) is \( a(t) = \frac{dv}{dt} = 2c - 6bt \). This shows acceleration also varies with time and is influenced by constants \( c \) and \( b \).
- **The sign of acceleration** is crucial because it tells us whether the particle is speeding up or slowing down. If \( a(t) > 0 \), the particle is accelerating, and if \( a(t) < 0 \), it is decelerating.
- For example, at \( t = 1 \text{ s} \), substituting into the acceleration equation gives \( a(1) = 2c - 6b(1) = -6 \text{ m/s}^2 \). Such calculations reveal when and how the motion dynamics change.

Displacement

Displacement refers to the change in position of the particle along the x-axis from the starting point to an ending point, not considering the actual path taken. It is simply the difference between the initial and final positions.
- In this exercise, displacement from \( t = 0 \text{ s} \) to \( t = 4 \text{ s} \) is calculated by finding \( x(t) \) at these times and subtracting. Using the given equation, \( x(0) = 0 \) and \( x(4) = 3(4)^2 - 2(4)^3 = -80 \text{ meters} \).
- Therefore, the displacement is \( x(t=4) - x(t=0) = -80 - 0 = -80 \text{ meters} \). This tells us the net change in position, which can be positive or negative depending on whether the particle has moved to the right or left of the starting point.
- Displacement is **not necessarily** the total distance traveled, as it’s possible for a particle to move forward and backward during its course.

Motion along the X-axis

Motion along the x-axis involves analyzing how the particle moves back and forth in a straight line as time progresses. This is a typical example of one-dimensional motion.
- It involves understanding concepts like the position of the particle at any given time, which can be determined using an equation like \( x(t) = ct^2 - bt^3 \).
- The motion's dynamics become clear by checking the velocity and acceleration equations. By doing so, we detect how the motion shifts direction or accelerates/decelerates as time goes on.
- Checking position, velocity, and acceleration at various points allows us to understand the complete motion narrative within the given time interval. For example, checking when the velocity is zero can show when the particle reaches its maximum position or changes direction.
- This type of motion is fundamental as it helps form the basics of learning about dynamics in physics, facilitating an understanding of more complex multidimensional motion scenarios.