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A \(2540-\mathrm{kg}\) test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by \(v(t)=\) \(A t+B t^{2},\) where \(A\) and \(B\) are constants and time is measured from the instant the fuel is ignited. At the instant of ignition, the rocket has an upward acceleration of 1.50 \(\mathrm{m} / \mathrm{s}^{2}\) and 1.00 s later an upward velocity of 2.00 \(\mathrm{m} / \mathrm{s}\) (a) Determine \(A\) and \(B\) , including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket's weight. (d) What was the initial thrust due to the fuel?

Step by step solution

01

Express Initial Conditions

From the problem, we know two conditions involving acceleration and velocity.1. Initial acceleration when the rocket starts, time \( t = 0 \), is \( 1.50 \, \mathrm{m/s^2} \).2. Velocity at \( t = 1.00 \, \mathrm{s} \) is \( 2.00 \, \mathrm{m/s} \).

02

Differentiate to Find Acceleration

Velocity is given by \( v(t) = At + Bt^2 \). To find acceleration, differentiate \( v(t) \) with respect to \( t \):\[ a(t) = \frac{dv}{dt} = A + 2Bt. \]At \( t=0 \), set \( a(0) = 1.50 \, \mathrm{m/s^2} \) to find \( A \,:\)\[ 1.50 = A.\]

03

Solve for Constant B

With \( v(1) = 2 \, \mathrm{m/s} \), substitute into the velocity equation:\[ v(1) = A \cdot 1 + B \cdot 1^2 = 2. \]We have \( A = 1.50 \) from Step 2, so:\[ 1.50 + B = 2. \]Thus, \( B = 0.50 \, \mathrm{m/s^3}. \)

04

Calculate Acceleration at 4 Seconds

Use the derivative \( a(t) = A + 2Bt \) with \( t = 4 \) to find acceleration at that time:\[ a(4) = 1.50 + 2 \times 0.50 \times 4 = 1.50 + 4.00 = 5.50 \, \mathrm{m/s^2}.\]

05

Calculate Thrust Force at 4 Seconds

Apply the formula for net force \( F_{net} = ma \) where \( m = 2540 \, \mathrm{kg} \) and \( a = 5.50 \, \mathrm{m/s^2} \):\[ F_{net} = 2540 \times 5.50 = 13970 \, \mathrm{N}. \]Thrust force \( F_{thrust} = F_{net} + mg \) where \( g = 9.81 \, \mathrm{m/s^2} \):\[ F_{thrust} = 13970 + 2540 \times 9.81 = 38889.4 \, \mathrm{N}. \]

06

Calculate Initial Thrust Force

For initial thrust at \( t = 0 \), use initial acceleration \( a = 1.50 \, \mathrm{m/s^2} \):\[ F_{net\,\, at\,\, start} = 2540 \times 1.50 = 3810 \, \mathrm{N}. \]Include gravity force to find initial thrust:\[ F_{thrust\,\, at\,\, start} = 3810 + 2540 \times 9.81 = 28375.4 \, \mathrm{N}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thrust Force Calculation

Understanding how to calculate the thrust force of a rocket is crucial in physics, especially when studying rocket motion equations. To find the thrust force, we need to consider both the net force produced by the rocket engine and the gravitational force acting on the rocket. This is expressed as: \( F_{\text{thrust}} = F_{\text{net}} + mg \), where \( m \) is the mass of the rocket and \( g \) is the acceleration due to gravity (\(9.81 \, \mathrm{m/s^2}\)).
Net force \( F_{\text{net}} \) is calculated using Newton's second law, \( F_{\text{net}} = ma \), where \( a \) is the rocket's acceleration at a given time.

• For initial conditions, the force calculation considers the rocket's initial upward acceleration.
• At any time, thrust force accounts for the acceleration at that particular time, as shown for the 4-second calculation in the exercise.

Calculating thrust force helps determine how efficiently a rocket can launch and make adjustments for better performance.

Rocket Acceleration

Rocket acceleration refers to how the speed of a rocket changes over time, which is derived from the given velocity function. In this context, the velocity function is \( v(t) = At + Bt^2 \).
To find the acceleration, differentiate the velocity function with respect to time \(t\). This gives us the expression \( a(t) = A + 2Bt \).

• This expression tells us that rocket acceleration depends on both constants \( A \) and \( B \), which are determined using initial conditions provided in the problem.
• By substituting the known values at specific times, such as \( t = 0 \) for initial acceleration, we can solve for these constants effectively.

Through careful calculation, rocket acceleration provides valuable insights into the rocket's performance at different times during its flight.

Differential Equations in Physics

Differential equations play a significant role in modeling physical phenomena, like rocket motion. They help describe how the velocity, acceleration, and other variables change over time.
The velocity equation \( v(t) = At + Bt^2 \) is an example of a time-dependent function derived from differential equations. Differentiating this equation with respect to time gives us the acceleration formula \( a(t) = A + 2Bt \).

• Solving these differential equations allows us to understand and predict the rocket's behavior over time.
• They provide a systematic approach to finding unknown variables such as constants \( A \) and \( B \), based on known physical conditions.

By applying differential equations, one can model real-world scenarios far beyond rocket science, making them essential in physics and engineering.