Problem 34 - Calculus Early Transcendentals

Question

A car slows down with an acceleration of \(a(t)=-15 \mathrm{ft} / \mathrm{s}^{2} .\) Assume that \(v(0)=60 \mathrm{ft} / \mathrm{s}\) and \(s(0)=0\). a. Determine and graph the position function, for \(t \geq 0\) b. How far does the car travel in the time it takes to come to rest?

Step-by-Step Solution

Answer

#Answer: The position function s(t) is -7.5t^2 + 60t, and the car travels 120 ft before it comes to a stop.

Step 1: Find the Velocity Function

To find the velocity function, we must integrate the acceleration function, \(a(t)=-15 ft/s^2\). $$v(t) = \int a(t) dt = \int -15 dt$$ Now integrate: $$v(t) = -15t + C$$ We are given the initial velocity \(v(0)=60 ft/s\). Use this to find the value of the integration constant C: $$60 = -15(0) + C$$ So, $$C = 60$$ Hence, the velocity function is: $$v(t) = -15t + 60$$

Step 2: Find the Position Function

Now, we need to find the position function by integrating the velocity function: $$s(t) = \int v(t) dt = \int (-15t + 60) dt$$ Now integrate: $$s(t) = -\frac{15}{2}t^2 + 60t + D$$ We are given the initial position \(s(0)=0\). Use this to find the value of the integration constant D: $$0 = -\frac{15}{2}(0)^2 + 60(0) + D$$ So, $$D = 0$$ Hence, the position function is: $$s(t) = -\frac{15}{2}t^2 + 60t$$

Step 3: Graph the Position Function

To graph the position function \(s(t)=-\frac{15}{2}t^2 + 60t\), we must plot it for \(t\geq0\). This is a quadratic function with a negative leading coefficient, so it will have the shape of an inverted parabola. The vertex of the parabola occurs at the maximum point for the car's position, which is determined by finding the derivative of the function and setting it equal to 0.

Step 4: Find the Time it Takes to Come to Rest

To find the time it takes for the car to come to rest, we must determine when the car's velocity becomes 0. Set \(v(t)\) equal to 0 and solve for \(t\): $$0 = -15t + 60$$ $$t = \frac{60}{15}$$ $$t = 4s$$ So the car comes to rest at \(t=4s\).

Step 5: Calculate the Distance Traveled Before the Car Stops

Now, we will use the position function to find how far the car travels in the time it takes to come to rest. We found that the car comes to rest at \(t=4s\), so plug this into the position function: $$s(4) = -\frac{15}{2}(4)^2 + 60(4)$$ $$s(4) = -\frac{15}{2}(16) + (240)$$ $$s(4) = -120 + 240$$ $$s(4) = 120 ft$$ So, the car travels a distance of 120 ft in the 4 seconds it takes to come to a stop.

Key Concepts

Understanding Acceleration

Acceleration is a measure of how quickly the velocity of an object changes over time. In this problem, the acceleration is specified as \(a(t) = -15 \, \text{ft/s}^2\). This negative value indicates that the car is slowing down — it’s "decelerating." The constant acceleration implies that the rate of change of velocity is steady throughout the observed time period.

• Positive acceleration increases velocity.
• Negative acceleration (deceleration) decreases velocity.
• Zero acceleration means the velocity is constant.

An interesting fact about acceleration in calculus is that it is the second derivative of the position function with respect to time. This relationship highlights how acceleration controls the curvature of the position graph and affects its overall shape.

Exploring Velocity

Velocity is the speed of an object in a specific direction. Unlike speed, it has a directional component. When we talk about velocity in this calculus problem, we are directly interested in how fast and in which direction the car is moving.
To find the velocity function from a given acceleration, we integrate the acceleration over time. Starting with \(a(t) = -15 \, \text{ft/s}^2\), integration gives the velocity function:\[v(t) = -15t + C\]Given that \(v(0) = 60 \, \text{ft/s}\), we solve for the constant \(C\) to get \(C = 60\). Thus, the velocity function becomes:\[v(t) = -15t + 60\]
This formula tells us that the car's velocity is decreasing linearly over time due to the constant negative acceleration. When the velocity hits zero, the car stops moving. By setting \(v(t) = 0\), we find the time \(t\) when the car comes to rest.

Determining the Position Function

The position function provides a detailed narrative of the car’s journey by telling us its location at any given time. We find it by integrating the velocity function. From the velocity function \(v(t) = -15t + 60\), integration gives:\[s(t) = -\frac{15}{2}t^2 + 60t + D\]Where \(D\) is the constant of integration. Using the initial condition \(s(0) = 0\), we determine \(D = 0\). Consequently, the position function is:\[s(t) = -\frac{15}{2}t^2 + 60t\]
This quadratic formula describes how the car's position changes over time; it's a parabolic curve that first rises, reaching a maximum point before it starts to descend. The car's movement can be tracked explicitly by plugging various values of \(t\) into this equation to observe how the position changes.

Integration in Calculus

Integration is one of the fundamental techniques in calculus. It's the act of finding a function whose derivative matches a given function. In real-world terms, integration often helps us determine accumulative quantities like areas, volumes, positions, etc.

• To find velocity from acceleration, we integrate the acceleration function.
• To find position from velocity, we integrate the velocity function.
• The constants added during integration embody initial conditions or starting states.

In this exercise, integration transforms acceleration into velocity and velocity into position. This chain of integrations illustrates the layered relationship between these terms, letting us move through "layers" of motion — from abstract acceleration to concrete position over time. Moreover, integration reflects areas under curves, offering a visual and intuitive comprehension of how functions accumulatively relate.