Problem 226 - Calculus Volume 1

Question

Suppose that a particle moves along a straight line with acceleration defined by \(a(t)=t-3, \quad\) where \(0 \leq t \leq 6\) (in meters per second). Find the velocity and displacement at time \(t\) and the total distance traveled up to \(t=6\) if \(v(0)=3\) and \(d(0)=0\)

Step-by-Step Solution

Answer

Velocity: \(v(t) = \frac{t^2}{2} - 3t + 3\); Displacement: \(d(t) = \frac{t^3}{6} - \frac{3t^2}{2} + 3t\); Total distance: 6 meters.

Step 1: Find the Velocity Function

To find velocity, integrate the acceleration function. Given that acceleration is defined by \( a(t) = t - 3 \), we integrate this function with respect to \( t \). \[ v(t) = \int (t - 3) \, dt = \frac{t^2}{2} - 3t + C \]We use the initial condition \( v(0) = 3 \) to find \( C \). Substituting \( t = 0 \) and \( v(0) = 3 \):\[ 3 = \frac{0^2}{2} - 3(0) + C \Rightarrow C = 3 \]Thus, the velocity function is:\[ v(t) = \frac{t^2}{2} - 3t + 3 \]

Step 2: Find the Displacement Function

To find the displacement, integrate the velocity function. The velocity is given by \( v(t) = \frac{t^2}{2} - 3t + 3 \). Integrate this function with respect to \( t \): \[ d(t) = \int \left( \frac{t^2}{2} - 3t + 3 \right) dt \]\[ d(t) = \frac{t^3}{6} - \frac{3t^2}{2} + 3t + D \]Using the initial condition \( d(0) = 0 \), we find \( D \):\[ 0 = \frac{0^3}{6} - \frac{3(0)^2}{2} + 3(0) + D \Rightarrow D = 0 \]Thus, the displacement function is:\[ d(t) = \frac{t^3}{6} - \frac{3t^2}{2} + 3t \]

Step 3: Calculate Total Distance Traveled by t=6

To find the total distance traveled, we evaluate the displacement function from \( t=0 \) to \( t=6 \). We need to find \( d(6) \):\[ d(6) = \frac{6^3}{6} - \frac{3(6)^2}{2} + 3(6) \]\[ d(6) = 36 - 54 + 18 \]\[ d(6) = 0 \]Since velocity changes direction at some points in the interval and the displacement is zero, we need to verify the intervals where the velocity is positive to accumulate the total distance:1. Solve \( v(t) = 0 \) to find turning points: \[ \frac{t^2}{2} - 3t + 3 = 0 \] Solving yields roots at \( t = 1 \) and \( t = 3 \).2. Calculate distances for intervals: - From \( t=0 \) to \( t=1 \), and \( t=3 \) to \( t=6 \) the direction is consistently forward. - Individual displacements can thus be computed as sum of absolute values:The total distance is obtained from calculating \( |d(1)| + |d(3) - d(1)| + |d(6) - d(3)| = 3 + 0 + 3 = 6 \).

Key Concepts

Integration

Integration is a core concept in calculus that allows us to find functions from their derivatives. It is essentially the reverse process of differentiation. In the context of motion, integration is used to derive formulas like velocity from acceleration.

• When we have an acceleration function, integrating it gives us the velocity function.
• Similarly, integrating the velocity function provides the displacement function.

Integration is often symbolized by the integral sign \( \int \), and it involves summing up infinitely small quantities to get a total. To find the constant of integration, we often use initial conditions provided in a problem.

Velocity Function

The velocity function describes how fast an object is moving and in which direction, at any point in time. It is derived from the acceleration function through integration. Here, the process involved integrating the function \( a(t) = t - 3 \) to obtain the velocity function \( v(t) = \frac{t^2}{2} - 3t + 3 \).
Some critical points regarding the velocity function include:

• Initial conditions, such as \( v(0) = 3 \), are used to determine any constants during integration.
• The velocity function can change signs, indicating changes in the direction of motion.

To solve or predict motion characteristics accurately, understanding how the velocity function is derived and behaves is essential.

Displacement

Displacement refers to the overall change in position of a particle or object. It is not the same as the distance traveled since it considers direction. To find displacement, we integrate the velocity function, which in this example is \( v(t) = \frac{t^2}{2} - 3t + 3 \).
By integrating the velocity, we get the displacement function: \( d(t) = \frac{t^3}{6} - \frac{3t^2}{2} + 3t \). Here are a few key points to remember:

• Displacement depends on the direction of motion; hence it can be negative, positive, or zero.
• We apply initial conditions, like \( d(0) = 0 \), to solve for constants when integrating.

The displacement can guide in understanding where a particle is relative to its starting position.

Particle Motion

Particle motion involves studying how particles move and change position over time. This includes understanding how velocity and acceleration impact their trajectory.
When analyzing particle motion:

• Acceleration tells us how the velocity of a particle changes.
• From velocity, we can calculate displacement over time.

In solving such problems, comprehending changes in motion directions is crucial to accurately calculate total distance traveled, which can be different from net displacement. In our example, although the overall displacement by \( t=6 \) was zero, the total distance traveled was determined to be 6 meters by considering intervals where the velocity changes direction.