Problem 27 - Calculus Early Transcendentals

Question

During the first \(40 \mathrm{s}\) of a rocket flight, the rocket is propelled straight up so that in \(t\) seconds it reaches a height of \(s=0.3 t^{3} \mathrm{ft}\). (a) How high does the rocket travel in \(40 \mathrm{s}\) ? (b) What is the average velocity of the rocket during the first \(40 \mathrm{s}\) ? (c) What is the average velocity of the rocket during the first \(1000 \mathrm{ft}\) of instight? (d) What is the instantancous velocity of the rocket at the end of \(40 \mathrm{s} ?\)

Step-by-Step Solution

Answer

(a) 19200 ft;(b) 480 ft/s;(c) ≈69.35 ft/s;(d) 1440 ft/s.

Step 1: Determine height at 40 seconds

We are given the height function: \( s = 0.3t^3 \). To find the height of the rocket at \( t = 40 \) seconds, substitute \( t = 40 \) into the equation: \[ s = 0.3 (40)^3 = 0.3 \times 64000 = 19200 \text{ ft} \]. Thus, the height of the rocket at 40 seconds is 19200 feet.

Step 2: Calculate average velocity for 40 seconds

The average velocity over a time interval can be determined using the formula: \[ \text{Average velocity} = \frac{\Delta s}{\Delta t} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \]For the interval from \( t = 0 \) to \( t = 40 \):\[ \text{Average velocity} = \frac{s(40) - s(0)}{40 - 0} = \frac{19200 - 0}{40} = 480 \text{ ft/s} \]. Thus, the average velocity during the first 40 seconds is 480 ft/s.

Step 3: Determine time to reach 1000 feet

First, solve for time \( t \) when the height \( s = 1000 \text{ ft} \) using the height equation:\( 0.3t^3 = 1000 \).Divide both sides by 0.3: \( t^3 = \frac{1000}{0.3} \approx 3333.33 \).Take the cube root: \( t \approx \sqrt[3]{3333.33} \approx 14.42 \text{ seconds} \).

Step 4: Calculate average velocity for first 1000 feet

Now, use the average velocity formula for the interval from \( t = 0 \) to \( t \approx 14.42 \):\[ \text{Average velocity} = \frac{1000 - 0}{14.42 - 0} \approx 69.35 \text{ ft/s} \]. Thus, the average velocity during the first 1000 feet is approximately 69.35 ft/s.

Step 5: Calculate instantaneous velocity at 40 seconds

To find the instantaneous velocity, determine the derivative of \( s = 0.3t^3 \) with respect to \( t \), which is the velocity function: \[ v(t) = \frac{d}{dt}(0.3t^3) = 0.9t^2 \].Now, substitute \( t = 40 \) into \( v(t) \):\[ v(40) = 0.9 \times (40)^2 = 0.9 \times 1600 = 1440 \text{ ft/s} \]. The instantaneous velocity of the rocket at 40 seconds is 1440 ft/s.

Key Concepts

average velocity

To find the average velocity of an object, we consider its change in position over a certain period of time. This gives us an idea of how fast the object is moving on average across that time span.
For example, if we know the starting and ending heights of a rocket and the time it takes to travel between these heights, we can use the formula: \[ \text{Average velocity} = \frac{\Delta s}{\Delta t} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \]
This formula takes the change in height \(\Delta s\) and divides it by the change in time \(\Delta t\).

• Average velocity gives a simple overview of motion over a period.
• It is affected by the total path traveled, not specific points in time.

In our example, for the rocket's trip during the first 40 seconds, we used this formula with the rocket starting at 0 feet and reaching 19200 feet in 40 seconds, giving an average velocity of 480 ft/s.

instantaneous velocity

Instantaneous velocity, unlike average velocity, tells us how fast an object is moving at an exact point in time.
This is like checking the speedometer of a car at an exact moment rather than calculating the average speed over a trip.
To find instantaneous velocity, we need to use calculus, specifically derivatives.
For the rocket, we computed its instantaneous velocity at exactly 40 seconds using the derivative of the height function, \( s(t) \), with respect to time \( t \).

• Instantaneous velocity provides a precise speed at a given moment.
• It requires the derivative of the position function.

In our rocket's case, we found the derivative of its height function \( s = 0.3t^3 \), which lead us to the velocity function \( v(t) = 0.9t^2 \). Evaluating this at \( t = 40 \) seconds showed us that the rocket's instantaneous velocity was 1440 ft/s.

derivatives

Derivatives are a fundamental concept in calculus that help us understand how functions change.
They can be thought of as measuring the slope of a function at any given point. In practical terms, they tell us the rate at which one quantity changes in relation to another.
For the rocket, its height function \( s(t) = 0.3t^3 \) gives us the derivative \( v(t) = 0.9t^2 \), representing the velocity of the rocket as it moves over time.

• Derivatives can determine rates of change like velocity and acceleration.
• They help us zero in on an object's speed at any moment.

With the derivative, we switched from a position function to a velocity function, making it possible to calculate both average and instantaneous velocity effectively.

height function

A height function in physics represents how an object's elevation changes over time.
For our rocket, the height function is given by \( s(t) = 0.3t^3 \), indicating that as time progresses, the height increases in proportion to \( t^3 \).
This cubed relationship shows that the rocket's height increases faster over time as it continues to rise.

• Height functions are crucial for tracking vertical motion.
• Ensure calculations are made for both heights and velocities.

Using the height function, we calculated the rocket's height at any given time, including at 40 seconds for modeling how far it has traveled vertically. Understanding the height function allows us to explore both how high it gets and how quickly it accelerates.