Problem 27 - 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. $$a(t)=-9.8, v(0)=20, s(0)=0$$

Step-by-Step Solution

Answer

Question: Determine the position and velocity functions of an object given its acceleration function, initial velocity, and initial position. Acceleration function: $$a(t) = -9.8$$ Initial velocity: $$v(0) = 20$$ Initial position: $$s(0) = 0$$ Position function: $$s(t) = -4.9t^2 + 20t$$ Velocity function: $$v(t) = -9.8t + 20$$

Step 1: Integrate the acceleration function

To find the velocity function, integrate the acceleration function with respect to time: $$v(t) = \int a(t) dt = \int -9.8 dt.$$

Step 2: Apply the initial velocity

To find the constant of integration, use the initial velocity, $$v(0) = 20.$$ Thus, integrate: $$v(t) = -9.8t + C,$$ and substitute the initial velocity: $$20 = -9.8(0) + C.$$ Solving for C, we find: $$C = 20.$$ So the velocity function is: $$v(t) = -9.8t + 20.$$

Step 3: Integrate the velocity function

To find the position function, integrate the velocity function with respect to time: $$s(t) = \int v(t) dt = \int (-9.8t + 20)dt.$$

Step 4: Apply the initial position

To find the constant of integration, use the initial position, $$s(0) = 0.$$ Thus, integrate: $$s(t) = -4.9t^2 + 20t + D,$$ and substitute the initial position: $$0 = -4.9(0)^2 + 20(0) + D.$$ Solving for D, we find: $$D = 0.$$ So the position function is: $$s(t) = -4.9t^2 + 20t.$$ Therefore, the position function of the object is $$s(t) = -4.9t^2+20t$$ and the velocity function is $$v(t)=-9.8t+20$$.

Key Concepts

Acceleration

Acceleration is the rate at which an object's velocity changes over time. It tells us how quickly an object is speeding up or slowing down.
In physics, acceleration can be positive or negative, depending on whether the object is gaining speed or losing speed, respectively. For example, in this problem, the given acceleration is \(a(t) = -9.8\) m/s².
This negative value indicates that the object is decelerating or slowing down at a constant rate. This is typical for objects under the influence of gravity near Earth's surface. When working with acceleration, it's crucial to understand that it is a vector quantity. This means it has both a magnitude and a direction. The direction of the acceleration vector will determine whether the object speeds up or slows down in its motion.

• Positive Value: Indicates an increase in speed.
• Negative Value: Suggests a decrease in speed or deceleration.

With acceleration, you can determine the velocity and position of an object using calculus, as demonstrated in this exercise.

Velocity

Velocity describes how fast an object is moving in a particular direction. It is a vector quantity, which means it includes both speed and direction.
In this exercise, we started with the constant acceleration function \(a(t) = -9.8\).
To determine the velocity, we integrated the acceleration function with respect to time. Integration allowed us to transition from the rate of change of velocity, back to the velocity expression itself.The equation for velocity after integration is \(v(t) = -9.8t + C\), where \(C\) is the integration constant.
By applying the initial condition \(v(0) = 20\), we determined that \(C = 20\). Thus, the velocity function becomes \(v(t) = -9.8t + 20\).
This tells us that the object starts with a velocity of 20 m/s and its speed decreases by 9.8 m/s every second due to acceleration.

• Initial Velocity: The speed at which an object begins its journey, here \(v(0) = 20\) m/s.
• Variable Term: Indicates how the velocity changes over time due to acceleration.

Understanding velocity is crucial in predicting how and where an object will move over time.

Integration

Integration is a fundamental concept in calculus, used to find functions such as velocity and position from a given rate of change, like acceleration.
It reverses differentiation, allowing us to accumulate effects over time.
In this exercise, integration helps us determine the velocity function from acceleration and the position function from velocity.The first integration is from acceleration to velocity. By integrating \(a(t) = -9.8\) with respect to time, we get the velocity function \(v(t) = -9.8t + 20\).
Here, we must apply initial conditions to solve for constants of integration, ensuring our function accurately represents the physical situation.Next, we integrate the velocity function \(v(t) = -9.8t + 20\) to find the position function.
The result is \(s(t) = -4.9t^2 + 20t + D\), and by applying the condition \(s(0) = 0\) m, we find \(D = 0\).
This gives us the position function \(s(t) = -4.9t^2 + 20t\), describing the object's location at any given time.

• Fundamental Process: Converting rates of change into cumulative distance and velocity.
• Integration Constants: Derived from initial conditions to match the physics problem.

Mastering integration is essential for solving real-world motion problems, like predicting an object's future position and velocity.