Problem 29 - 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. $$a(t)=-0.01 t, v(0)=10, s(0)=0$$

Step-by-Step Solution

Answer

Question: Determine the position and velocity functions of an object moving along a straight line with an acceleration function given by $$a(t) = -0.01t$$, an initial velocity of $$v(0) = 10$$, and an initial position of $$s(0) = 0$$. Answer: The position function is $$s(t) = -\frac{1}{3}(0.005)t^3 + 10t$$, and the velocity function is $$v(t) = -0.005t^2 + 10$$.

Step 1: Integrate Acceleration Function to Find Velocity Function

First, we need to integrate the given acceleration function with respect to time: $$a(t) = -0.01 t$$ $$\int a(t) \, dt = \int(-0.01 t) \, dt$$ $$v(t) = -0.005t^2 + C_1$$ Where C_1 is the integration constant. To find C_1, we'll use the initial velocity given, v(0) = 10: $$v(0) = -0.005(0)^2 + C_1 = 10 \Rightarrow C_1 = 10$$ Thus, the velocity function is: $$v(t) = -0.005t^2 + 10$$

Step 2: Integrate Velocity Function to Find Position Function

Now, we need to integrate the velocity function with respect to time: $$v(t) = -0.005t^2 + 10$$ $$\int v(t) \, dt = \int(-0.005t^2 + 10) \, dt$$ $$s(t) = -\frac{1}{3} * 0.005t^3 + 10t + C_2$$ Where C_2 is the integration constant. To find C_2, we'll use the initial position given, s(0) = 0: $$s(0) = -\frac{1}{3} * 0.005(0)^3 + 10(0) + C_2 = 0 \Rightarrow C_2 = 0$$ Thus, the position function is: $$s(t) = -\frac{1}{3}(0.005)t^3 + 10t$$ In conclusion, the position function is $$s(t) = -\frac{1}{3}(0.005)t^3 + 10t$$, and the velocity function is $$v(t) = -0.005t^2 + 10$$.

Key Concepts

Acceleration function

The acceleration function describes how an object's acceleration changes over time. In this exercise, we have an acceleration function given by \( a(t) = -0.01t \). This function indicates that the acceleration is not constant but decreases linearly as time \( t \) progresses. The negative sign shows that it acts in the opposite direction of motion. Such a function can be interpreted as a deceleration effect.Understanding the effects of acceleration functions is crucial in physics and engineering because it helps predict future velocity and position state. To move from acceleration to velocity, you need to integrate the acceleration function with respect to time. This process requires you to find the antiderivative or indefinite integral of the acceleration function.In the initial step of solving the problem, the integration of the acceleration provides the velocity function, as explained in the mathematical operations conducted.

Velocity function

The velocity function represents the rate at which an object's position changes over time. In the given problem, after integrating the acceleration function \( a(t) = -0.01t \), we end up with a velocity function \( v(t) = -0.005t^2 + C_1 \).To determine the constant \( C_1 \), we use the initial velocity, which is given as \( v(0) = 10 \). Substituting into the velocity equation helps us solve for \( C_1 \), yielding \( C_1 = 10 \). Therefore, the complete velocity function becomes:

• \( v(t) = -0.005t^2 + 10 \)

This function tells us that:

• The velocity decreases as time increases because of the \( -0.005t^2 \) term.
• The object starts with a velocity of 10 units at \( t = 0 \).

Understanding velocity in terms of its function is significant since it serves as a bridge to finding the object's position over time.

Position function

The position function describes the specific location of an object in relation to a reference point and is obtained by integrating the velocity function. In our exercise, the velocity function \( v(t) = -0.005t^2 + 10 \) was integrated to find the position function \( s(t) \).Upon integrating the velocity function, we arrive at:

• \( s(t) = -\frac{1}{3}(0.005)t^3 + 10t + C_2 \)

To determine the constant \( C_2 \), we use the initial position \( s(0) = 0 \). This implies that at the start, the object was at the origin, or reference point. Substituting back allows us to solve for \( C_2 \), which in this case is zero. Hence, the final position function simplifies to:

• \( s(t) = -\frac{1}{3}(0.005)t^3 + 10t \)

This function explains:

• How the object's position changes over time due to its velocity's quadratic and linear components.
• The impact of time on displacement, highlighting the effects of decreasing velocity.

By understanding this function, you can predict the object's future positions and the path it has traveled along the line.