91Ó°ÊÓ

Thrust is the force exerted by a rocket's engines that propels the rocket forward. In our exercise, it's crucial to understand how to calculate thrust at various stages of the rocket launch. The thrust calculation relies on Newton's second law of motion, which states that force equals mass times acceleration (\( F = ma \)). In this problem, we specifically calculate thrust when the rocket is 4 seconds into its flight.
To find the thrust force at this point, we first determine the rocket's acceleration at \( t = 4 \, ext{s} \). With a rocket mass of \( 2540 \, ext{kg} \) and acceleration of \( 5.50 \, ext{m/s}^2 \), the calculation becomes:

• \( F_{\text{thrust}} = 2540 \, \text{kg} \times 5.50 \, \text{m/s}^2 = 13970 \, \text{N} \)

This thrust not only propels the rocket but also counteracts the gravitational force acting on it. Calculating thrust at each stage of the rocket's journey helps engineers understand performance and adjust as needed.

The vertical velocity function describes how fast a rocket moves upward over time. In this exercise, the given function is \( v(t) = At + Bt^2 \), with \( A \) and \( B \) being constants that we must determine. This function demonstrates the rocket's increase in speed as it climbs vertically.

To find \( A \) and \( B \), we use initial conditions given in the problem: an upward acceleration of \( 1.50 \, ext{m/s}^2 \) at ignition (\( t = 0 \)) and an upward velocity of \( 2.00 \, ext{m/s} \) after \( 1 \, ext{s} \). By differentiating the velocity function to find acceleration, we set up these equations:

• \( a(t) = \frac{dv}{dt} = A + 2Bt \)
• At \( t = 0 \), \( a(0) = A \), so \( A = 1.50 \, ext{m/s}^2 \)
• At \( t = 1 \, ext{s} \), \( 1.50 + B = 2.00 \), so \( B = 0.50 \, ext{m/s}^2 \)

Understanding this function allows us to predict the rocket’s performance over time and is essential for planning its trajectory.

Newton's Second Law is pivotal in understanding rocket propulsion. This law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration (\( F = ma \)). In the context of a rocket, this tells us how much force is needed to propel it upwards against the force of gravity.
When considering the initial launch, the required thrust must do more than just accelerate the rocket. It must also overcome gravitational force. The overall force required can be calculated by summing the force needed for both, such as:

• Initial net force for upward acceleration: \( F_{\text{net}} = m \times a(0) = 2540 \, \text{kg} \times 1.50 \, \text{m/s}^2 = 3810 \, \text{N} \)
• Add gravitational force \( (W = mg = 2540 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 24911.4 \, \text{N}) \).
• Initial total thrust force \( F_{\text{thrust, initial}} = 3810 \, \text{N} + 24911.4 \, \text{N} = 28721.4 \, \text{N} \).

This conceptual basis is critical for engineers to design engines capable of lifting rockets off the ground efficiently.

Instantaneous acceleration is the rate at which an object's velocity changes at a specific moment in time. In this exercise, it helps us understand how rapidly the rocket accelerates at any given second after ignition. The derivative of the velocity function gives us the instantaneous acceleration function, \( a(t) = A + 2Bt \).

For this particular rocket, we calculated constant \( A \) and \( B \) previously. To find the acceleration at a specific time, such as 4 seconds after ignition, substitute \( t = 4 \) into the acceleration function:

• \( a(4) = 1.50 + 2 \times 0.50 \times 4 = 5.50 \, \text{m/s}^2 \)

Knowing instantaneous acceleration at various points helps manage the rocket’s fuel use and optimize its path through space, ensuring it follows the intended trajectory safely and efficiently.