Problem 37 - Calculus Early Transcendentals

Question

At \(t=0,\) a train approaching a station begins decelerating from a speed of \(80 \mathrm{mi} / \mathrm{hr}\) according to the acceleration function \(a(t)=-1280(1+8 t)^{-3},\) where \(t \geq 0\) is measured in hours. How far does the train travel between \(t=0\) and \(t=0.2 ?\) Between \(t=0.2\) and \(t=0.4 ?\) The units of acceleration are \(\mathrm{mi} / \mathrm{hr}^{2}\)

Step-by-Step Solution

Answer

Question: Calculate the distance the train travels between \(t=0\) and \(t=0.2\) and between \(t=0.2\) and \(t=0.4\), given the acceleration function \(a(t) = -1280(1+8t)^{-3}\) and an initial velocity of \(80\ \mathrm{mi/hr}\).

Step 1: Find the velocity function

To find the velocity function, integrate the acceleration function with respect to time and add a constant of integration, which represents the initial velocity. $$v(t) = \int a(t) dt + C = \int{-1280(1+8t)^{-3}} dt + C$$ Integrate the given acceleration function, and then substitute the initial velocity to find the constant of integration.

Step 2: Finding the constant of integration

To find the constant of integration, use the initial condition, which states that at \(t=0\), the velocity is \(80\ \mathrm{mi/hr}\). $$v(0) = 80\ \mathrm{mi/hr} = C$$

Step 3: Find the position function

Next, integrate the velocity function with respect to time to find the position function. $$s(t) = \int v(t) dt + K = \int{-1280(1+8t)^{-3}+80} dt + K$$ This gives the position function of the train.

Step 4: Calculate the distance between t=0 and t=0.2

Now, evaluate the position function at \(t=0\) and \(t=0.2\). Find the difference between these two values to calculate the distance travelled. $$\Delta s = s(0.2) - s(0)$$

Step 5: Calculate the distance between t=0.2 and t=0.4

Similarly, evaluate the position function at \(t=0.2\) and \(t=0.4\), and find the difference between these two values to calculate the distance travelled. $$\Delta s = s(0.4) - s(0.2)$$

Key Concepts

Integrating Acceleration Function

To understand how far an object has traveled over time, we start with its acceleration function, which tells us how the object's velocity is changing. In calculus, integrating the acceleration function gives us the object's velocity function.

This process involves finding the antiderivative of the acceleration function. If the acceleration is given by a(t), then the velocity function v(t) is obtained by the integral of a(t) with respect to time t. The general form of this integral is:
\[v(t) = \text{the integral of } a(t) \, dt + C\]
where C is the constant of integration. This constant corresponds to the initial velocity of the object, which is required to solve for a specific velocity function. Without this initial condition, there could be infinitely many velocity functions corresponding to the acceleration function.

Initial Velocity in Calculus

The initial velocity plays a critical role when deriving the velocity function from acceleration. It acts as a starting point for velocity and is denoted by v(0) or v_0.

In our train example, the initial velocity is given as 80 mi/hr at time t = 0. This value is essential because it determines the constant of integration C when we integrate the acceleration function.

To integrate correctly, we use the fact that the velocity is defined by this initial value at time zero by stating v(0) = 80. This equality allows us to solve for C, thus obtaining a complete velocity function that accurately represents the train's motion from the start of deceleration.

Position Function Calculus

In calculus, the position function represents an object's location over time. Once we have the velocity function, we can find the position function by integrating the velocity function with respect to time. This is similar to how we found the velocity function from the acceleration.

Mathematically, if v(t) is the velocity function, then the position function s(t) is given by:

Integration of Velocity to Position

\[s(t) = \text{the integral of } v(t) \, dt + K\]
Here, K is another constant of integration. To solve for K, we need an initial position, often given as s(0). In the absence of an initial position, the constant K can be considered zero, assuming the initial position is at the origin. However, with a specific initial position, we can adjust K accordingly to reflect the object's actual starting point.

Calculus Distance Calculation

Calculating the distance an object has traveled within a particular time interval involves evaluating the position function at the start and end of the interval. The difference between these two values gives us the distance traveled.

For example, when determining how far the train traveled between t=0 and t=0.2, we evaluate the position function s(t) at both time points and subtract:

Distance Evaluation

\[\Delta s = s(0.2) - s(0)\]
This calculation reflects the net distance the train covered in the given time frame. Similarly, to find out how far the train traveled between t=0.2 and t=0.4 hours, we would calculate:

Subsequent Distance Evaluation

\[\Delta s = s(0.4) - s(0.2)\]
By using these methods, we can derive meaningful interpretations of distance from the train's position function over its time of travel.