Problem 61 - 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. \( a(t) = t + 4 \), \( v(0) = 5 \), \( 0 \le t \le 10 \)

Step-by-Step Solution

Answer

(a) Velocity: \( v(t) = \frac{t^2}{2} + 4t + 5 \). (b) Distance: 416.67 meters.

Step 1: Understand the Problem

We are given the acceleration function \( a(t) = t + 4 \) and the initial velocity \( v(0) = 5 \). We need to find the velocity function \( v(t) \) and the distance traveled over the interval \( 0 \leq t \leq 10 \).

Step 2: Find the Velocity Function

To find the velocity function \( v(t) \), integrate the acceleration function \( a(t) = t + 4 \). The indefinite integral of \( t + 4 \) is \( \frac{t^2}{2} + 4t + C \), where \( C \) is the integration constant.

Step 3: Determine the Integration Constant

Use the initial condition \( v(0) = 5 \) to find \( C \). Set \( v(0) = \frac{0^2}{2} + 4 \times 0 + C = 5 \), so \( C = 5 \). Therefore, the velocity function is \( v(t) = \frac{t^2}{2} + 4t + 5 \).

Step 4: Find the Distance Function

To find the distance traveled, calculate the definite integral of the velocity function \( v(t) \) from 0 to 10. The distance function \( s(t) \) is the integral of \( v(t) = \frac{t^2}{2} + 4t + 5 \).

Step 5: Integrate the Velocity Function

The integral of \( \frac{t^2}{2} + 4t + 5 \) is \( \frac{t^3}{6} + 2t^2 + 5t \). Evaluate it from 0 to 10 to find the distance traveled: \[s(10) - s(0) = \left(\frac{10^3}{6} + 2 \times 10^2 + 5 \times 10\right) - 0\].

Step 6: Calculate the Distance Traveled

Calculate the value: \( \frac{1000}{6} + 200 + 50 = \frac{1000}{6} + 250 \). \( \frac{1000}{6} = 166.67 \). Therefore, the distance traveled is \( 416.67 \) meters.

Key Concepts

Acceleration Function

Understanding how objects move requires us to consider acceleration. The acceleration function, often expressed as \( a(t) \), represents how the velocity of an object changes over time. In this problem, the acceleration function is given by \( a(t) = t + 4 \). This tells us that as time increases, our acceleration does too, starting at 4 units when \( t = 0 \) and increasing by 1 unit for each second that passes. By observing this function, we can see that the particle is accelerating faster each second.

To transition from acceleration to velocity, integration is necessary. Acceleration gives us the rate of change of velocity, and understanding this is key to predicting the future state of moving objects. Remember that without acceleration (i.e., it's constant), velocity would remain unchanged.

Integration in Calculus

Integration is a fundamental concept in calculus used to find areas under curves or accumulate quantities, such as velocity from acceleration. When given an acceleration function, you integrate to find the velocity function. By integrating \( a(t) = t + 4 \), we obtain the velocity function \( v(t) = \frac{t^2}{2} + 4t + C \).

Here, the process transforms acceleration (rate of change) into velocity, which is a cumulative total of how fast something is moving at a given time. Exploring limits through definte integration helps us compute exact distances or time intervals. Additionally, an integration constant \( C \) usually appears, which needs initial condition to solve. Think of \( C \) as extra information about the system, which can be calculated with parameters like initial velocity.

Initial Velocity

Initial velocity is the starting speed of an object before any forces like acceleration start acting on it. It's represented by \( v(0) \) when at time zero. In this exercise, the initial velocity \( v(0) = 5 \) m/s implies that the particle is already moving when observations begin.

Finding the integration constant \( C \) is a crucial step. The given initial velocity provides the necessary boundary condition—by substituting \( t=0 \) into the integrated velocity function, we obtain \( C = 5 \). This gives us a complete velocity function: \( v(t) = \frac{t^2}{2} + 4t + 5 \).

The importance of initial velocity lies in its maintenance of continuity and accuracy in predictive modeling. Without this value, it would result in uncertainty when attempting to complete the time-dependent equation for velocity.