Problem 32 - Calculus Early Transcendentals

Question

Find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity, and initial position. $$a(t)=\frac{20}{(t+2)^{2}}, v(0)=20, s(0)=10$$

Step-by-Step Solution

Answer

Question: Determine the position and velocity functions for an object moving along a straight line with an acceleration function of $$a(t) = \frac{20}{(t+2)^{2}}$$, initial velocity of 20 m/s, and initial position of 10 m. Answer: The position function of the object is given by $$s(t) = -20\ln|t+2| + 30t + 10 + 20\ln{2}$$, and the velocity function is given by $$v(t) = -\frac{20}{t+2} + 30$$.

Step 1: Integrate the acceleration function to find the velocity function

To find the velocity function, we need to integrate the acceleration function with respect to time. The acceleration function is given as $$a(t) = \frac{20}{(t+2)^{2}}$$. By integrating with respect to time, we obtain the result: $$\int a(t) dt = v(t) = \int \frac{20}{(t+2)^2}dt$$ Now we need to find the antiderivative of the acceleration function. We use substitution method. Let $$u = t + 2$$ and $$du = dt$$. $$v(t) = \int \frac{20}{u^2} du = -\frac{20}{u} + C_1 = -\frac{20}{t+2} + C_1$$ Next, we use initial velocity condition to find $$C_1$$. The initial velocity is given as: $$v(0) = 20 = -\frac{20}{0+2} + C_1$$ Solving for $$C_1$$, we get $$C_1 = 30$$. Therefore, the velocity function is: $$v(t) = -\frac{20}{t+2} + 30$$

Step 2: Integrate the velocity function to find the position function

Next, we need to find the position function, which is done by integrating the velocity function with respect to time: $$\int v(t) dt = s(t) = \int \left(-\frac{20}{t+2} + 30\right) dt$$ Now, we separate the integral into two different integrals: $$s(t) = \int\left(-\frac{20}{t+2}\right) dt + \int 30 dt$$ Integrating both functions individually and combining their results: $$s(t) = -20 \ln|t+2| + 30t + C_2$$ Now, we use the initial position condition to find $$C_2$$. The initial position is given as: $$s(0) = 10 = -20\ln{(0+2)} + 30(0) + C_2$$ Solving for $$C_2$$, we get $$C_2 = 10 + 20\ln{2}$$. Therefore, the position function is: $$s(t) = -20\ln|t+2| + 30t + 10 + 20\ln{2}$$ The position function s(t) and the velocity function v(t) of the object moving along the straight line are given by: $$s(t) = -20\ln|t+2| + 30t + 10 + 20\ln{2}$$ and $$v(t) = -\frac{20}{t+2} + 30$$

Key Concepts

Integrating the Acceleration Function

When an object moves along a straight line, its acceleration at any given moment can describe how quickly it is speeding up or slowing down. To find out how this acceleration affects the velocity of the object over time, we integrate the acceleration function with respect to time.

Integrating an acceleration function like \( a(t) = \frac{20}{(t+2)^2} \) essentially involves finding the area under the acceleration-time curve from a starting point to time \( t \). This process gives us the velocity function minus the constant of integration. Using mathematical methods such as substitution can simplify the integration process. In our exercise, the result after integration and using the initial conditions is the velocity function \( v(t) = -\frac{20}{t+2} + 30 \).

This step uncovers how the object's speed changes over time and is critical for understanding its movement.

Finding the Velocity Function

The velocity function is directly obtained by integrating the acceleration function. It reveals at what speed and in which direction the object is moving at any point in time. For analytical purposes, recognizing that the velocity is the first derivative of the position function and the integral of the acceleration function is essential. By integrating the specific acceleration function given in the problem, we derive the object’s velocity.

It is through this calculated velocity function that we can explain the object's motion and predict its future speed. Moreover, applying the initial velocity, we find the particular solution to the problem. So, the velocity function isn't just a formula, it presents a narrative of the object's journey over time.

Initial Velocity and Position

Understanding initial conditions is crucial in problems that involve motion. The initial velocity, represented as \( v(0) \), and the initial position, symbolized as \( s(0) \), serve as starting points in our calculations.

When we integrate the acceleration function to find velocity, or velocity to find position, we always get a generic function plus a constant denoted as \( C \). This constant \( C \) can only be determined by applying the initial conditions provided in the problem. Intuitively, you can think of it as calibrating your results to match the specific scenario of the object’s motion at the start of observation. In our example, the initial conditions led to the constants \( C_1 = 30 \) and \( C_2 = 10 + 20\ln{2} \), fitting the velocity and position functions to the object's particular situation.

Position Function Calculus

The position function is the final step in understanding the entire picture of an object's movement. By integrating the velocity function with respect to time, we get the position function, which tells us the location of the object at any time \( t \).

Calculus comes into play once more as we perform this integration, looking out for any constants that emerge, and then apply our initial conditions to determine the specific position function. The formula \( s(t) = -20\ln|t+2| + 30t + 10 + 20\ln{2} \) gives the precise position of our object at any selected time. With this function in hand, you can visualize the path and position of the object as it moves along the straight line, serving as the roadmap of its journey.