Problem 35 - 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

Answer: The train travels approximately 6.4516 miles between t=0 and t=0.2, and 1.6129 miles between t=0.2 and t=0.4.

Step 1: Finding the velocity function

To find the velocity function, we need to integrate the acceleration function with respect to time t. Let's integrate the acceleration function: $$ v(t) = \int a(t) dt = \int -1280(1+8 t)^{-3} dt $$

Step 2: Solving the integral

We can use substitution to solve this integral. Let \(u = 1 + 8t\) and \(du = 8 dt\). We now have: $$ v(t) = \int -1280 u^{-3} \frac{1}{8} du = -160\int u^{-3} du $$ Now integrate \(u^{-3}\) and substitute back for u: $$ v(t) = -160[\frac{u^{-2}}{-2}] + C = 80u^{-2} + C = 80(1 + 8t)^{-2} + C $$

Step 3: Finding the constant C

We can find the constant C by noting that at \(t=0\), the initial speed of the train is given as \(80mph\). Therefore: $$ v(0) = 80(1 + 0)^{-2} + C = 80 \Rightarrow C = 0 $$ So the velocity function is: $$ v(t) = 80(1 + 8t)^{-2} $$

Step 4: Finding the position function

To find the position function, we need to integrate the velocity function with respect to time t: $$ s(t) = \int v(t) dt = \int 80(1 + 8t)^{-2} dt $$

Step 5: Solving the integral

Again, use substitution. Let \(u' = 1+8t\) and \(du' = 8 dt\). We get: $$ s(t) = \int 80u'^{-2}\frac{1}{8} du' = 10\int u'^{-2} du' $$ Now integrate \(u'^{-2}\) and substitute back for u': $$ s(t) = 10[\frac{u'^{-1}}{-1}] + C' = -10u'^{-1} + C' = -10(1 + 8t)^{-1} + C' $$

Step 6: Finding the constant C'

At \(t=0\), we can assume the position to be 0 for simplicity. Therefore, $$ s(0) = -10(1 + 0)^{-1} + C' = 0 \Rightarrow C' = 10 $$ So the position function is: $$ s(t) = -10(1 + 8t)^{-1} + 10 $$

Step 7: Evaluating the position function

We need to find the distance traveled between \(t = 0\) and \(t = 0.2\) and between \(t = 0.2\) and \(t = 0.4\). Plug these values into the position function. For \(t=0.2\): $$ s(0.2) = -10(1 + 8(0.2))^{-1} + 10 \approx 6.4516 $$ For \(t=0.4\): $$ s(0.4) = -10(1 + 8(0.4))^{-1} + 10 \approx 4.8387 $$

Step 8: Calculating distances

Finally, we calculate the distances: - Between \(t = 0\) and \(t = 0.2\) : \(s(0.2) - s(0) = 6.4516 - 0 = 6.4516\) miles - Between \(t = 0.2\) and \(t = 0.4\) : \(s(0.4) - s(0.2) = 4.8387 - 6.4516 = -1.6129\) miles Since the distance cannot be negative, we take the absolute value of the second result. Therefore, the train travels approximately: - 6.4516 miles between \(t=0\) and \(t=0.2\) - 1.6129 miles between \(t=0.2\) and \(t=0.4\)

Key Concepts

Integration

Integration is a powerful mathematical tool that helps us find quantities like velocity and position from their rates of change, such as acceleration. When we integrate an acceleration function with respect to time, we're essentially accumulating all the subtle changes in velocity over a given time interval. Integration simplifies working with continuous change in motion, showing how small incremental changes add up to a bigger picture.

For example, given an acceleration function, the process of integration helps us find the velocity function. This is done by integrating the acceleration function over time, which mathematically is expressed as:

• \[ v(t) = \int a(t) \, dt \]

Here, \( v(t) \) represents the velocity as a function of time. This step is crucial in moving from understanding how fast something is speeding up or slowing down, to knowing its precise speed at any given moment during its motion.

Velocity function

The velocity function describes how fast an object is moving and in what direction. Deriving it requires integrating the acceleration function, as acceleration is the rate of change of velocity.

Consider a scenario where the train's acceleration function is given as:

• \[ a(t) = -1280(1+8t)^{-3}\]

We integrated this function and found the velocity function:

• \[ v(t) = 80(1 + 8t)^{-2}\]

This velocity function reflects how the train's speed changes over time. Importantly, it also incorporates initial conditions (like starting speed), which help determine the specific form of the function. By setting \( t = 0 \) and knowing the initial velocity, we solve for any constants that crop up during integration. These constants ensure our function accurately represents real-world motion.

Position function

The position function gives the location of an object at a specific time. It's the integral of the velocity function; by accumulating velocities, we determine how far an object has traveled.

For our train deceleration problem, once the velocity function was established as:

• \[ v(t) = 80(1 + 8t)^{-2}\]

We integrated it further to find the position function:

• \[ s(t) = -10(1 + 8t)^{-1} + 10\]

This equation reveals the train's position over time, crucial for determining how far it covers certain intervals. For practical use, such functions provide insights into distances over specified times by simply plugging in different values of \( t \). Initial conditions, like the starting position, allow us to solve and remove any resultant constants from our integration process, solidifying the function's accuracy.

Acceleration function

The acceleration function describes how the velocity of an object changes with time. In kinematics, it reflects the fundamental cause of changes in motion, dictating both speed up and slow down processes.

In the given scenario, the train's acceleration function was presented as:

• \[ a(t) = -1280(1+8t)^{-3}\]

This function, where acceleration is negative, indicates the train is decelerating. The distinct form \((1+8t)^{-3}\) shapes exactly how quickly the train is slowing down; here, the negative sign marks a direction opposite to movement, typical of deceleration.

Understanding this function is crucial, as it serves as the starting point for deriving both velocity and position functions, through integration. Its formulation provides insight into really how motion equations are interconnected: changes in acceleration not only influence speed directly but cascade into the positional shifts over time.