Problem 83 - Calculus of a Single Variable

Question

Motion Along a Line In Exercises \(81-84,\) the function \(s(t)\) describes the motion of a particle along a line. For each function, (a) find the velocity function of the particle at any time \(t \geq 0,\) (b) identify the time interval(s) in which the particle is moving in a positive direction, (c) identify the time intervale in which the particle is moving in a negative direction, and (d) identify the time(s) at which the particle changes direction. $$ s(t)=t^{3}-5 t^{2}+4 t $$

Step-by-Step Solution

Answer

The velocity function is \(v(t) = 3t^{2} - 10t + 4\). The particle moves in the positive direction when \(v(t) > 0\) and in the negative direction when \(v(t) < 0\). The particle changes its direction at the time points when \(v(t) = 0\).

Step 1: Determination of velocity function

The velocity function \(v(t)\) can be found by taking the derivative of \(s(t)\). So,\[v(t) = s'(t) = 3t^{2} - 10t + 4\].

Step 2: Identification of time intervals for positive direction

The particle moves in the positive direction when \(v(t) > 0\). To find the interval(s) for which the particle is moving in a positive direction, we need to solve the inequality \(3t^{2} - 10t + 4 > 0\). The solution of this inequality is the required interval of positive motion.

Step 3: Identification of time intervals for negative direction

The particle moves in the negative direction when \(v(t) < 0\). To find the interval(s) for which the particle is moving in a negative direction, we need to solve the inequality \(3t^{2} - 10t + 4 < 0\). The solution to this inequality is the required interval of negative motion.

Step 4: Identify times when particle changes direction

The particle changes direction when \(v(t)=0\). Set \(v(t) = 0\) and solve for \(t\) to find these time points.

Key Concepts

Derivative

In calculus, the derivative represents how a function changes as its input changes. Essentially, it gives you the rate of change or the slope of the function at any point. For the motion of a particle along a line, the derivative of the position function, usually denoted as \( s(t) \), is critical because it provides the particle's velocity.

Taking the derivative involves applying differentiation rules. For a polynomial like \( s(t) = t^3 - 5t^2 + 4t \), you differentiate each term individually:

• The derivative of \( t^3 \) is \( 3t^2 \).
• The derivative of \( -5t^2 \) is \( -10t \).
• The derivative of \( 4t \) is \( 4 \).

Combining these, the velocity function \( v(t) \) is the derivative of \( s(t) \): \[ v(t) = 3t^2 - 10t + 4 \] This formulation is crucial as it goes on to help determine other motion characteristics of the particle.

Velocity

Velocity is a measure of how fast something is moving and in which direction. In our exercise, velocity is the derivative of the position function, \( v(t) = 3t^2 - 10t + 4 \).

Velocity can tell us if a particle is moving forward or backward along a line:

• If \( v(t) > 0 \), the particle is moving in a positive direction (forward).
• If \( v(t) < 0 \), the particle is moving in a negative direction (backward).

Understanding the sign of the velocity function over time is key to determining the movement behavior of the particle.

Inequality

In the context of motion, solving inequalities helps us identify intervals of time where certain conditions are satisfied such as when a particle is moving forward or backward. For a particle moving along a line, we use inequalities to separate these motion intervals.

To find where \( v(t) > 0 \), solve the inequality:\[ 3t^2 - 10t + 4 > 0 \] And, to find where \( v(t) < 0 \), solve:\[ 3t^2 - 10t + 4 < 0 \] Solving these inequalities often involves using techniques like:

• Finding roots of the related equation \( v(t) = 0 \).
• Using test intervals or a sign chart to determine the intervals where the inequality holds true.

Understanding how to solve these inequalities is crucial for accurately determining the direction of motion.

Motion Along a Line

The concept of motion along a line concerns how a particle's position changes over time in one dimension. This change is expressed through the position function \( s(t) \), and its derivative gives us the velocity \( v(t) \), which points out how fast and in which direction the particle is moving.

To further understand the motion:

• Determine when the particle changes direction, which occurs when \( v(t) = 0 \) as the velocity transitions from positive to negative or vice versa.
• Analyze the solution of \( v(t) = 0 \) for roots. These are the critical points where the particle might change direction.
• Review intervals determined from inequalities to sketch a complete overview of the particle’s motion.

Comprehending motion along a line requires an integration of these analyses to visualize how the particle behaves across time intervals.