Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

A certain rocket, initially at rest, is shot straight up with an acceleration of \(6 t\) meters per second per second during the first 10 seconds after blast- off, after which the engine cuts out and the rocket is subject only to gravitational acceleration of \(-10 \mathrm{me}\) 5ors per second per second. How high will the rocket go?

Step-by-Step Solution

Answer

The rocket reaches a maximum height of 6000 meters.

Step 1: Determine acceleration function

Initially, the rocket's acceleration is given as a function: \( a(t) = 6t \). This is valid for the first 10 seconds. After 10 seconds, the acceleration is due to gravity: \( a(t) = -10 \), since gravity acts in the opposite direction of the initial acceleration.

Step 2: Integrate acceleration to find velocity

For \( t \leq 10 \), integrate the acceleration \( a(t) = 6t \) with respect to time to get the velocity: \( v(t) = \int 6t \, dt = 3t^2 + C \). At \( t = 0 \), initial velocity \( v(0) = 0 \), so \( C = 0 \). Thus, \( v(t) = 3t^2 \) for \( 0 \leq t \leq 10 \). For \( t > 10 \), the velocity changes due to gravity. We set \( v(10) = 300 \) since \( 3 imes 10^2 = 300 \). Use \( a(t) = -10 \) now to find the new velocity function: \( v(t) = 300 - 10(t - 10) = 400 - 10t \) for \( t > 10 \).

Step 3: Integrate velocity to find position

For \( t \leq 10 \), integrate the velocity \( v(t) = 3t^2 \) to find position: \( s(t) = \int 3t^2 \, dt = t^3 + C \). With initial condition \( s(0) = 0 \), we find \( C = 0 \). So, \( s(t) = t^3 \) for \( 0 \leq t \leq 10 \). Compute \( s(10) = 10^3 = 1000 \). For \( t > 10 \), integrate the velocity \( v(t) = 400 - 10t \): \( s(t) = 400t - 5t^2 + C_1 \). Using the continuity of position at \( t = 10 \), set \( s(10) = 1000 \): \( 1000 = 4000 - 5(10)^2 + C_1 \). Hence, \( C_1 = 1000 - 3000= -2000 \).

Step 4: Find maximum height

To find maximum height, determine when velocity becomes zero after engine cutoff (i.e., for \( t > 10 \)). Set \( v(t) = 400 - 10t = 0 \), yielding \( t = 40 \). Calculate \( s(40) = 400(40) - 5(40)^2 - 2000 \) which simplifies to \( 16000 - 8000 - 2000 = 6000 \). Thus, the rocket reaches a maximum height of 6000 meters.

Key Concepts

Acceleration Function

In the context of rocket motion, understanding the acceleration function is crucial. For our specific problem, the rocket's acceleration function is dynamic. Initially, the rocket experiences a time-dependent acceleration given by the function \( a(t) = 6t \) during the first 10 seconds after takeoff. This implies that the rocket's acceleration is increasing linearly with time, making it go faster as each second passes.
After the initial 10 seconds, the rocket's engine cuts off, leaving the rocket subjected to the constant acceleration due to gravity, which is \( a(t) = -10 \) m/s². This negative sign indicates gravity acts in the opposite direction of the rocket's initial motion.
By understanding these distinct phases of acceleration, we can predict the rocket's motion and behavior over time.

Velocity Integration

Once we have the acceleration function, the next step is integrating it to determine the rocket's velocity over time. For the initial phase, where \( t \leq 10 \), the acceleration is \( a(t) = 6t \). Integrating this function, we use the process of finding the antiderivative, which gives us the velocity function \( v(t) = 3t^2 + C \).
Given the rocket starts from rest, the initial velocity \( v(0) = 0 \), which means the constant \( C = 0 \), resulting in the velocity function \( v(t) = 3t^2 \). This reflects how the velocity increases quadratically during the powered flight phase.
After the engine shuts off at \( t = 10 \) seconds, we account for gravitational acceleration with \( a(t) = -10 \). By integrating this, considering the velocity at \( t = 10 \) to be 300 m/s, the new velocity function becomes \( v(t) = 400 - 10t \), showing how the velocity decreases linearly as gravity pulls the rocket back down.

Gravitational Acceleration

Gravitational acceleration plays a significant role after the rocket's engine cuts out. Specifically, it is characterized by the constant value of \(-10\) m/s². This acceleration is crucial because it continuously affects the rocket, decelerating it during the ascent and then accelerating it on its downward return.
Knowing the gravitational acceleration helps us calculate how long the rocket can ascend past the engine cutoff. The constant force of gravity opposes the initial upward velocity, reducing it to zero before causing a descent.
From this point onward, integration helps us find how the position changes over time given the known acceleration due to gravity. This understanding allows us to determine both the peak height the rocket achieves and the duration of its flight.

Maximum Height Calculation

To determine the maximum height of the rocket, we need to know when the upward velocity becomes zero after the engine cuts off, at which point it will stop rising and begin falling back down. This occurs when we set the post-10 second velocity function \( v(t) = 400 - 10t \) to zero, resulting in \( t = 40 \).
Once we have this time, we compute the position function \( s(t) \) for \( t > 10 \) to find the exact height. The function \( s(t) = 400t - 5t^2 - 2000 \) is derived from integrating the velocity function. Substituting \( t = 40 \), we perform the calculations to find the maximum height of the rocket:

• Calculate \( 400 \times 40 = 16000 \)
• Calculate \( 5 \times 40^2 = 8000 \)
• Add and subtract appropriately to get \( 16000 - 8000 - 2000 = 6000 \)

Thus, the maximum height is 6000 meters, demonstrating the peak of the rocket's trajectory.