Question

The position of a particle along the \(x\) axis depends on the time according to the equation \(x=A t^{2}-B t^{3}\), where \(x\) is in meters and \(t\) is in seconds. (a) What SI units must \(A\) and \(B\) have? For the following, let their numerical values in SI units be \(3.0\) and \(1.0\), respectively. (b) At what time does the particle reach its maximum positive \(x\) position? ( \(c\) ) What total path-length does the particle cover in the first 4 seconds? ( \(d\) ) What is its displacement during the first 4 seconds? ( \(e\) ) What is the particle's velocity at the end of each of the first 4 seconds? \((f)\) What is the particle's acceleration at the end of each of the first 4 seconds? \((g)\) What is the average velocity for the time interval \(t=2\) to \(t=4 \mathrm{~s}\) ?

Short answer

Based on the steps, the units of \(A\) and \(B\) are \(m/s^2\) and \(m/s^3\) respectively. The time of maximum \(x\) is \(2A/3B\). The total path length and displacement as well as the values of velocity and acceleration at the end of each of the first 4 seconds and the average velocity between \(t=2\) and \(t=4\) are all solvable by substituting \(A=3\) and \(B=1\) into their respective equations and solving numerically.

Step-by-step solution

Units of A and B Values

Since \(x\) is in metres and \(t\) is in seconds. The SI units for \(A\) and \(B\) can be determined by equating the units of \(x\) with the units in the equation. Hence, the unit for \(A\) is \(m/s^2\) and for \(B\) is \(m/s^3\)

Time of Maximum X

The time of maximum \(x\) can be found by differentiating \(x\) and setting it equal to 0. This gives \(2At-3Bt^2=0\). Factoring out \(t\), we find \(t=2A/3B\).

Total Path Length and Displacement

To find both the total path length and displacement, we integrate the speed (absolute of the derivative of \(x\)) from \(t=0\) to \(t=4\) and the integrate the derivative of \(x\) from \(t=0\) to \(t=4\) respectively. This gives us the total path length as \(\int_0^4|2At-3Bt^2|dt\) and the displacement as \(\int_0^4 (2At-3Bt^2) dt\). To obtain numerical values, substitute \(A=3\) and \(B=1\) and solve.

Velocity

Velocity of the particle at the end of each of the first four seconds can be found by evaluating the expression for \(dx(t)/dt\) at \(t=1, 2, 3, 4\). In this case, \(v(t) = dx(t)/dt = 2At-3Bt^2\). Substitute \(A=3\) and \(B=1\) to get numerical values.

Acceleration

Acceleration of the particle can be found by differentiating the velocity expression. This gives \(a(t) = dv(t)/dt = 2A-6Bt\). Evaluate at \(t=1, 2, 3, 4\) for each of the first four seconds.

Average Velocity

The average velocity over the interval \([t=2,t=4]\) can be found by integrating the velocity function and then dividing by the time interval. This is given by \(v_{avg} = (1/2) \int_2^4 v(t) dt\), where \(v(t) = 2At-3Bt^2\)

Key concepts

SI units in physics

In physics, the International System of Units (SI) provides a standardized approach to measuring various physical properties. For instance, in kinematics, which deals with motion, the fundamental units for distance and time are meters (m) and seconds (s), respectively. When we encounter a kinematic equation like

\( x = A t^2 - B t^3 \)

in measurement of a particle's position, we must ensure that the coefficients 'A' and 'B' have appropriate units to maintain consistency. Here, the term 'A' must have units of meters per second squared (\( m/s^2 \)) to comply with the \( t^2 \) factor, and 'B' must have units of meters per second cubed (\( m/s^3 \)) to align with the \( t^3 \) term. This is crucial to understand since consistent units allow us to accurately describe and predict physical phenomena.

Motion along a straight line

When a particle moves along a straight line, its motion can be fully described by its position along that line over time. The mathematical representation of this position is a function of time, which allows us to track where the particle is, at any given moment. For example, in our exercise, the position function

\( x(t) = A t^2 - B t^3 \)

illustrates how the particle’s position changes with each passing second. The equation tells us that the particle’s motion is influenced by the squared and cubed terms of time, suggesting that its speed and direction vary as time passes. Concepts of kinematics, such as velocity and acceleration, help us unpack the nature of this motion further.

Velocity and acceleration

Velocity and acceleration are foundational concepts in kinematics, providing insights into the rate of change of an object's position and the rate of change of its velocity, respectively.

The velocity of an object is the derivative of its position with respect to time (\( v(t)= \frac{dx(t)}{dt} \) ) and represents both speed and direction of motion. Acceleration, in turn, is the derivative of velocity with respect to time (\( a(t) = \frac{dv(t)}{dt} \) ), highlighting how an object's velocity changes over time. These derivatives play a central role in analyzing and predicting the motion of objects.

Calculating Velocity

In the provided exercise, when we compute the velocity (\( v(t) \) ) for the given position function, we find it to be \( v(t) = 2At - 3Bt^2 \) . This expression allows us to determine the instantaneous velocity at any desired time by plugging in the corresponding value of \( t \).

Calculating Acceleration

To find the particle’s acceleration, we take the derivative of its velocity: \( a(t) = 2A - 6Bt \) . This formula signifies that the particle’s acceleration is not constant but dependent on time, providing a more complex understanding of its motion.

Path length and displacement

Path length and displacement are two different ways of measuring the extent of a particle's movement. Path length refers to the total distance traveled by the particle, regardless of its direction, making it a scalar quantity – a measurement that only has magnitude.

On the other hand, displacement is a vector quantity, meaning it includes both magnitude and direction. Displacement measures how far the particle is from its starting position and in which direction.

Computing Path Length

The path length in the given exercise can be obtained by integrating the absolute value of the velocity over time, which gives us the total distance traveled: \(\int_0^4 |2At - 3Bt^2| dt \) .

Computing Displacement

To find displacement, we integrate the velocity itself, which can result in positive or negative values, symbolizing direction: \(\int_0^4 (2At - 3Bt^2) dt \) . This distinction between path length and displacement is important, as it impacts how we interpret the motion of any object.

Time derivatives in physics

In physics, the concept of time derivatives is used to describe how a physical quantity changes with respect to time. Kinematics, in particular, relies heavily on derivatives to express velocity and acceleration, as seen in our example.

Derivatives stem from calculus and provide a mathematical way to determine the instantaneous rate of change. For instance, the first derivative of the position function with respect to time gives us the velocity, indicating how quickly position changes. Further, taking the derivative of velocity gives us acceleration, which shows how velocity changes over time.

These time derivatives are crucial tools for scientists and engineers; they not only predict future motion but also understand the influences affecting an object's trajectory. In kinematic problems like the one we're reviewing, derivatives allow us to answer critical questions about the behavior of moving objects.