Problem 62 - Calculus: Early Transcendentals

Question

The acceleration function (in \( m/s^2 \)) and the initial velocity are given for a particle moving along a line. Find (a) the velocity at time \( t \) and (b) the distance traveled during the given time interval. \( a(t) = 2t + 3 \), \( v(0) = -4 \), \( 0 \le t \le 3 \)

Step-by-Step Solution

Answer

Velocity at time \( t \): \( v(t) = t^2 + 3t - 4 \). Distance traveled: \( \frac{5}{2} \text{ meters} \).

Step 1: Write the Integral for Velocity

The velocity function, \( v(t) \), is found by integrating the acceleration function, \( a(t) = 2t + 3 \). Recall that the integral of acceleration gives velocity.

Step 2: Integrate the Acceleration Function

Integrate \( a(t) = 2t + 3 \):\[ v(t) = \int (2t + 3) \, dt = \int 2t \, dt + \int 3 \, dt \]Applying the power rule, we get:\[ v(t) = t^2 + 3t + C \]where \( C \) is the integration constant.

Step 3: Determine the Integration Constant

Use the initial condition \( v(0) = -4 \): Substitute \( t = 0 \) into the velocity equation:\[ v(0) = 0^2 + 3(0) + C = -4 \]This gives \( C = -4 \).

Step 4: Write the Expression for Velocity

Substitute \( C = -4 \) back into the velocity equation:\[ v(t) = t^2 + 3t - 4 \]

Step 5: Set Up the Integral for Distance

To find the distance traveled, integrate the velocity function over the interval \( [0, 3] \). This is given by the integral of the absolute value of velocity:\[ s(t) = \int_{0}^{3} |v(t)| \, dt \]

Step 6: Evaluate the Distance Integral

Since \( v(t) = t^2 + 3t - 4 \) needs to be non-negative within the interval for direct integration without using absolute value, solve for zeros of \( v(t) \):\[ t^2 + 3t - 4 = 0 \]Using the quadratic formula:\[ t = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm 5}{2} \]This gives \( t = 1 \) and \( t = -4 \). Only consider \( t = 1 \) since it is within the interval. Split the integral at \( t = 1 \):For \( 0 \le t < 1 \), \( v(t) < 0 \), and for \( 1 \le t \le 3 \), \( v(t) \geq 0 \), so:\[ s(t) = -\int_{0}^{1} v(t) \, dt + \int_{1}^{3} v(t) \, dt \]

Step 7: Integrate and Combine Results

Compute the two integrals:\[ \int_{0}^{1} (t^2 + 3t - 4) \, dt \] gives \( [-\frac{1}{3} + \frac{3}{2} - 4t]_{0}^{1} = -\frac{11}{6} \)\[ \int_{1}^{3} (t^2 + 3t - 4) \, dt \] gives \( \left[\frac{27}{3} + \frac{27}{2} - 12\right] - \left[-\frac{1}{3} + \frac{3}{2} - 4\right] = \frac{26}{3} \)Adding these:\[ -\left(-\frac{11}{6}\right) + \frac{26}{3} = \frac{5}{2} \]Thus, the total distance traveled is \( \frac{5}{2} \text{ meters} \).

Key Concepts

Understanding the Acceleration Function

In calculus, the acceleration function, denoted as \( a(t) \), describes how the velocity of an object changes over time. It's measured in units like \( m/s^2 \), which indicates the change in velocity in meters per second for every second. Understanding acceleration is crucial when analyzing the motion of objects, such as a particle moving along a line.
For example, if you have an acceleration function like \( a(t) = 2t + 3 \), it tells us that the acceleration increases linearly with time, meaning it becomes stronger as time progresses. The coefficient \( 2t \) means that the acceleration increases by \( 2 \) meters per second squared for each second that passes.
The constant \( 3 \) can be interpreted as the initial acceleration at \( t = 0 \). This combination of linear and constant components gives a comprehensive view of how the velocity will change over time.

The Velocity Function Explained

The velocity function, \( v(t) \), describes the speed of an object at a given time and is highly related to acceleration. To find the velocity from acceleration, we perform integration. The integral of the acceleration function gives us the velocity function, which shows the cumulative effect of acceleration over time.
Take the acceleration function \( a(t) = 2t + 3 \) as an example. To find the velocity, you would integrate this function:

• First, you integrate \( 2t \) using the power rule, resulting in \( t^2 \).
• Next, integrate the constant \( 3 \), which becomes \( 3t \).

Combined, these give us \( v(t) = t^2 + 3t + C \), where \( C \) is the constant of integration, determined by initial conditions. This constant is crucial as it adjusts the velocity function to match real-world data, like a known initial velocity. If given \( v(0) = -4 \), solving for \( C \) involves substituting \( t=0 \) in the equation, helping to finalize the velocity function: \( v(t) = t^2 + 3t - 4 \).

Integral of Velocity: Finding Distance

Calculating the distance an object travels involves integrating the velocity function. When you integrate velocity over time, you effectively sum up all small increments of distance, which results in the total distance traveled.
To find the distance when \( v(t) = t^2 + 3t - 4 \) between \( t = 0 \) and \( t = 3 \), you need to compute the integral of \( v(t) \). However, to ensure accuracy, you must monitor where the velocity is positive or negative because distance must be cumulative and always non-negative.

• If \( v(t) < 0 \) within an interval, reverse the sign of the integral.
• For intervals where \( v(t) \geq 0 \), simply integrate as is.

In this case, the quadratic formula reveals \( v(t) \) has a zero at \( t = 1 \). Thus, split the integration process:
From 0 to 1 where the velocity is negative becomes \(-\int_0^1 v(t) \, dt\), and from 1 to 3 where it is positive becomes \(\int_1^3 v(t) \, dt\). Summing up these calculated integrals results in the total distance the particle traveled, showing how integral calculus can be applied to understand real motion dynamics.