Problem 26 - 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. $$a(t)=-32, v(0)=50, s(0)=0$$

Step-by-Step Solution

Answer

Question: Determine the position and velocity functions for an object with an acceleration function $$a(t) = -32$$, an initial velocity of $$v(0) = 50$$, and an initial position of $$s(0) = 0$$. Answer: The position function for the object is $$s(t) = -16t^2 + 50t$$, and the velocity function is $$v(t) = -32t + 50$$.

Step 1: Integrate the acceleration function.

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

Step 2: Plug in the given acceleration function.

The acceleration function is given by $$a(t) = -32$$, so we will integrate this constant function with respect to time: $$v(t) = \int -32 \, dt$$

Step 3: Find the antiderivative.

The antiderivative of the constant function $$-32$$ is a linear function with a slope of $$-32$$: $$v(t) = -32t + C$$

Step 4: Apply the initial velocity condition.

We are given the initial velocity of the object, $$v(0) = 50$$, so we can find the constant $$C$$ by plugging in the initial condition: $$50 = -32(0) + C$$ $$C = 50$$

Step 5: Write the velocity function.

The velocity function is now found to be: $$v(t) = -32t + 50$$

Step 6: Integrate the velocity function.

To find the position function, we need to integrate the velocity function with respect to time: $$s(t) = \int v(t) \, dt$$

Step 7: Plug in the velocity function.

The velocity function we found is $$v(t) = -32t + 50$$, so we will integrate this linear function with respect to time: $$s(t) = \int (-32t + 50) \, dt$$

Step 8: Find the antiderivative.

The antiderivative of the linear function $$-32t + 50$$ is a quadratic function: $$s(t) = -16t^2 + 50t + D$$

Step 9: Apply the initial position condition.

We are given the initial position of the object, $$s(0) = 0$$, so we can find the constant $$D$$ by plugging in the initial condition: $$0 = -16(0)^2 + 50(0) + D$$ $$D = 0$$

Step 10: Write the position function.

Finally, the position function is found to be: $$s(t) = -16t^2 + 50t$$

Key Concepts

Acceleration

Acceleration is the rate of change of velocity over time. It tells us how quickly an object speeds up or slows down. In this problem, the acceleration is given as a constant value, specifically,

• \(a(t) = -32\)

This means that the object is experiencing a constant acceleration in the negative direction. Integrating this function would provide the velocity function, as acceleration is the derivative of velocity.

When integrating a constant acceleration,

• we essentially calculate the area under the horizontal line that represents this function.
• This helps us understand how the velocity changes with respect to time.

In general, the formula to obtain velocity from acceleration is:
\[v(t) = \int a(t)\,dt + C\] where \(C\) is the integration constant, found using initial conditions.

Velocity

Velocity is the speed and direction of an object's motion. Unlike speed, velocity is a vector quantity, meaning it has both magnitude and direction. In the exercise, the initial velocity is given as

• \(v(0) = 50\)

which indicates the starting speed of the object at time zero.

Knowing the constant acceleration, we integrated to find the velocity function:
\[v(t) = -32t + 50\]The integration constant, \(C\), was determined using the initial velocity condition. The velocity function is linear, indicating a consistent rate of change due to constant acceleration.

This function tells us:

• The initial upward velocity is 50 units per time interval.
• Every second, the velocity decreases by 32 units due to negative acceleration.

Position

Position refers to the location of an object at a particular time. It provides a reference point with respect to a starting position. In this exercise, the initial position is given by

• \(s(0) = 0\).

To find the position of the object as a function of time, we integrated the velocity function:
\[s(t) = -16t^2 + 50t\]This results from taking the antiderivative of the linear velocity function and applying the initial condition for position.

This position function is quadratic, reflecting a parabolic path common in motion with constant acceleration:

• \(-16t^2\) shows the effect of acceleration over time.
• \(50t\) represents the initial velocity's contribution to moving the object.

Using this function, one can calculate the object's position at any time \(t\). This provides a complete description of the object's motion along a line in terms of where it is, how fast it's moving, and how its velocity is changing.