Problem 31 - Calculus: Early Transcendentals

Question

Zero velocity A projectile is fired vertically upward and has a position given by \(s(t)=-16 t^{2}+128 t+192,\) for \(0 \leq t \leq 9\) a. Graph the position function, for \(0 \leq t \leq 9\) B. From the graph of the position function, identify the time at which the projectile has an instantancous velocity of zero; call this time \(t=a\) c. Confirm your answer to part (b) by making a table of average velocities to approximate the instantaneous velocity at \(t=a\) d. For what values of \(t\) on the interval [0,9] is the instantaneous velocity positive (the projectile moves upwand)? e. For what values of \(t\) on the interval [0,9] is the instantancous velocity negative (the projectile moves downward)?

Step-by-Step Solution

Answer

Answer: The projectile is moving upward when \(0 \leq t < 4\) and moving downward when \(4 < t \leq 9\).

Step 1: Graph the position function

First, we will graph the position function \(s(t)=-16 t^2+128 t + 192\) for \(0 \leq t \leq 9\). You can use any graphing calculator or software like Desmos or GeoGebra to do this. 

Step 2: Find the derivative to represent the instantaneous velocity

The position function is given by \(s(t)= -16t^2 + 128t + 192\). To find the instantaneous velocity, we will need to find its derivative with respect to time t: \(v(t) = \frac{ds}{dt} = -32t + 128\)

Step 3: Find when the projectile's instantaneous velocity is zero

To find when the projectile's instantaneous velocity is zero, we need to find the time t at which \(v(t) = 0\) \(-32t + 128 = 0 \Rightarrow t = 4\) The projectile's instantaneous velocity is zero at \(t=a=4\).

Step 4: Approximate instantaneous velocity at \(t=a\) using a table of average velocities

We can approximate the instantaneous velocity at \(t=a\) by making a table of average velocities around \(t=a\). For example, we can find average velocities at \(t=3.5, 3.9, 4.1, 4.5\): | t | v(t) | |-----|-------| | 3.5 | 16 | | 3.9 | 3.2 | | 4.1 | -3.2 | | 4.5 | -16 | As we can observe from the table, the average velocities transition from positive to negative at around \(t=a=4\). This confirms our answer in part (b).

Step 5: Determine when the instantaneous velocity is positive (projectile moves upward)

The instantaneous velocity is represented by the function \(v(t)=-32t + 128\). To find when it is positive, solve the inequality: \(-32t + 128 > 0 \Rightarrow t < 4\) The projectile moves upward when \(0 \leq t < 4\).

Step 6: Determine when the instantaneous velocity is negative (projectile moves downward)

Likewise, we can determine when the projectile moves downward by solving the inequality: \(-32t + 128 < 0 \Rightarrow t > 4\) The projectile moves downward when \(4 < t \leq 9\).

Key Concepts

Derivative of Position Function

To understand the motion of an object, it is essential to analyze its position function, particularly by obtaining the derivative of that function. In calculus, the derivative represents the rate of change. For a position function like s(t) = -16t^2 + 128t + 192, the derivative with respect to time t, noted as ds/dt or s'(t), gives us the instantaneous velocity of the object at any given time.

By differentiating the given function, we get v(t) = -32t + 128, which indicates the velocity of the object at time t. If the velocity is positive, the object is moving upwards; if it's negative, it's moving downwards. And if the velocity is zero, the object momentarily stops, having no movement in either direction.

Average Versus Instantaneous Velocity

Students often confuse average velocity with instantaneous velocity. Average velocity is defined for a time interval and is calculated by the total displacement divided by the total time taken. However, instantaneous velocity tells us how fast an object is moving at a particular instant.

To illustrate, consider calculating the average velocity of a projectile between two points in time. This gives an overall idea of its speed, but not its particular speed at any given moment within that span. In contrast, the instantaneous velocity—found through the derivative of the position function at a specific time—offers the precise rate at which the object's position is changing right at that moment.

Graphing Position Functions

The graph of a position function can provide a wealth of information about an object's motion. The function s(t) = -16t^2 + 128t + 192 represents the height of a projectile over time when plotted on a graph. At first, as t increases, s(t) also increases until the projectile reaches its peak height.

Using graphing tools like Desmos or GeoGebra, we can visualize the parabolic path typically followed by a projectile in a gravitational field. The graph assists in pinpointing key moments like when the object reaches its maximum height (where the graph reaches a vertex) or when it returns to the ground (where the graph intersects the t-axis).

Motion of a Projectile

The motion of a projectile fired vertically upwards is generally affected by gravity, which causes its upward motion to decelerate until it momentarily stops and starts to fall back down. For the function s(t) = -16t^2 + 128t + 192, graphing and differentiation are tools we can use to examine this motion.

The graph shows the ascent and descent in a symmetrical arc, where the top of the arc corresponds to an instantaneous velocity of zero—this is the peak of the projectile's flight. By looking at when the derivative of the position function is zero, we determine the time when the projectile is at its highest point (time t=a). Before this point, the velocity is positive (upwards movement), and after this point, the velocity becomes negative (downwards movement), demonstrating the effects of gravity on the projectile.