Problem 47 - Calculus Single Variable

Question

Concern an object that is propelled straight up. Its height at time \(t\) seconds is given in feet by \(H(t)=-16 t^{2}+128 t+68\). Compute the velocity and acceleration of the object.

Step-by-Step Solution

Answer

Velocity: \( v(t) = -32t + 128 \); Acceleration: \( a(t) = -32 \).

Step 1: Understanding the Problem

We are given a quadratic equation for the height of an object, where the height as a function of time is represented by \( H(t) = -16t^2 + 128t + 68 \). To find velocity and acceleration, we need to relate them to the derivatives of the height function with respect to time.

Step 2: Finding the Velocity Function

The velocity of the object \( v(t) \) is the first derivative of the height function \( H(t) \). By applying differentiation, we find: \[ v(t) = \frac{d}{dt}(-16t^2 + 128t + 68) = -32t + 128. \] This gives us the velocity function in terms of time.

Step 3: Finding the Acceleration Function

The acceleration of the object \( a(t) \) is the derivative of the velocity function \( v(t) \). Taking the derivative of \( v(t) = -32t + 128 \), we get: \[ a(t) = \frac{d}{dt}(-32t + 128) = -32. \] This means the acceleration is constant.

Key Concepts

Quadratic Functions

Quadratic functions are a type of polynomial function that are defined by a degree of two. This means you will see an equation in the form of \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. In the context of our exercise, the quadratic function is \( H(t) = -16t^2 + 128t + 68 \), which represents the height of an object over time.

One key feature of quadratic functions is their parabolic shape when graphed. The coefficient \( a \) determines whether the parabola opens upwards or downwards. In our case, \( a = -16 \), so the parabola opens downwards. This indicates that the object reaches a peak height and then falls back towards the ground.

Solving problems involving quadratic functions can involve finding roots or intercepts, the vertex (which gives the maximum or minimum point), and interpreting these within the context of the problem. Understanding the nature of the graph helps to visualize the motion of the object in this problem.

Velocity Calculation

Velocity is calculated as the rate of change of position with respect to time. For an object whose height is defined by the function \( H(t) \), its velocity \( v(t) \) is found by differentiating \( H(t) \) with respect to \( t \). This calls for some basic rules of differentiation.

In our example, we apply the rule of taking the first derivative to find the instantaneous rate of change in height, i.e., velocity:

• The derivative of \( -16t^2 \) is \( -32t \).
• The derivative of \( 128t \) is \( 128 \).
• The derivative of a constant \( 68 \) is \( 0 \).

Therefore, the velocity function is \( v(t) = -32t + 128 \). This provides a way to calculate the velocity at any time \( t \). Practical interpretation shows that as time passes, the term \( -32t \) becomes more negative, meaning that gravity slows down the object until it starts descending.

Acceleration Calculation

Acceleration is the rate of change of velocity over time. Much like finding velocity from height, acceleration can be found by differentiating the velocity function \( v(t) \). In our setting, it's crucial as it involves understanding how the velocity itself changes with each passing second.

From the derived velocity function \( v(t) = -32t + 128 \), we differentiate again to calculate the acceleration function:

• The derivative of \( -32t \) is \( -32 \).
• The derivative of constant \( 128 \) is \( 0 \).

Thus, the acceleration is a constant value \( a(t) = -32 \).

This constant acceleration indicates the object is uniformly affected by gravitational force, showing that irrespective of its velocity, the acceleration remains the same, simplifying predictions about the object's movement at any point in time. Overall, understanding acceleration aids in designing more efficient systems in physics and engineering.