/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 A \(2540 \mathrm{~kg}\) test roc... [FREE SOLUTION] | 91Ó°ÊÓ

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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 such 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. The rocket has an upward acceleration of \(1.50 \mathrm{~m} / \mathrm{s}^{2}\) at the instant of ignition 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?

Short Answer

Expert verified
The constants in the velocity equation are \(A=1.50 m/s\) and \(B=0.50 m/s²\). Four seconds after ignition, the rocket's acceleration is \(5.50 m/s²\), and the thrust exerted by the burning fuel is \(13970N\). The initial thrust exerted by the burning fuel is \(3810N\).

Step by step solution

01

Determine constants A and B

Given that the upward velocity of the rocket is given by the equation \(v(t)=A t + B t^{2}\) and the acceleration at the instant of ignition (t=0) is 1.50 m/s², we can deduce that \(A =1.50 m/s\). By substituting \(A \) in the velocity equation and given that at one seconds later (t=1s), the upward velocity is \( 2.0 m/s\), we can solve for \(B \) to get \(B= 0.50 m/s²\). Thus, \(A=1.50 m/s\) and \(B=0.50 m/s²\).
02

Find the acceleration at t = 4s

The acceleration \(a(t)\) can be obtained from the derivative of the velocity equation \(v(t)\). In this case, \(a(t) = A + 2Bt\). By substitizing \(A = 1.50 m/s\) and \(B = 0.50 m/s²\) and \(t = 4.00s\), we find that the acceleration at the given time is \(a(4) = 5.50 m/s²\).
03

Determine the force exerted by the burning fuel at t = 4s

Recall that according to Newton's second law, the net force \(F\) acting on an object is given by the product of its mass \(m\) and acceleration \(a\), i.e, \(F = m \cdot a\). By substitizing the rocket's mass \(m = 2540 kg\) and acceleration at \(t = 4s\) (\(a = 5.50 m/s²\)), we can compute the force exerted as \(F = 13970N\). This is the force that the burning fuel exerts on the rocket at t = 4s.
04

Calculate initial thrust

The initial thrust, i.e., thrust at ignition, is given by the force exerted by the fuel when the velocity is at its initial value. Since the acceleration at ignition is given as \(1.50 m/s²\), the initial thrust exerted by the burning fuel can be obtained from \(F = m \cdot a\) to be \(F_{initial} = 3810N\).

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Key Concepts

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

Kinematic Equations
When studying motion, kinematic equations are indispensable tools that employ initial velocity, constant acceleration, and time to describe an object's position and velocity at any given time. In the context of rocket physics, these equations relate the vertical velocity of the rocket to time (usually represented as v(t)). For a rocket with a velocity that can be modeled by the quadratic equation v(t) = At + Bt^2, the coefficients A and B derive from its initial acceleration and velocity conditions.

For example, to find the rocket's velocity at a particular moment, you would input the time into the equation. If you need to know the acceleration at any time, you calculate the derivative, which for our rocket gives a(t) = A + 2Bt. These kinematic relationships allow us to characterize the entire flight path of the rocket given initial conditions and constants such as A and B.
Thrust Force
The thrust force is what propels a rocket upwards and is a result of Newton’s third law of motion, which is often paraphrased as 'for every action, there is an equal and opposite reaction.' In the context of rockets, the action is the expulsion of gas out of the engine, and the reaction is the movement of the rocket in the opposite direction.

The thrust force comes from the burning of fuel, which generates high-speed gas that is expelled to produce thrust. It's important to note that this force can change over time as the fuel burns and as the rocket's weight changes due to fuel expenditure. This force is integral to overcoming gravity and any atmospheric resistance. When calculated at different time intervals, the thrust force provides insights into the rocket's performance and structural requirements during its ascent.
Newton's Second Law
Newton's second law is at the core of understanding rocket motion. It states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration (F = ma). In our rocket problem, we apply this law to calculate the thrust, given that the mass of the rocket is known and the acceleration can be derived from the kinematic equations, as we did when we calculated the initial thrust using the acceleration at ignition.

Furthermore, this law tells us that as the rocket's mass decreases due to the burning of fuel, for the same amount of thrust, the acceleration of the rocket will increase. This dynamic interplay between mass, force, and acceleration is fundamental to rocketry and allows missions to be planned and executed with precision.
Acceleration
Acceleration is the rate of change of velocity. It is a vector quantity, meaning it has both magnitude and direction. In our context, upward acceleration indicates the rocket is speeding up as it ascends, while a downward acceleration (or deceleration) would mean the rocket is slowing down. The initial acceleration is especially critical because it sets in motion the rocket's departure from the launch pad.

An interesting point to note is that the acceleration is not constant in the case of a rocket flight, unlike the simplified scenarios often encountered in basic physics problems. Due to varying thrust and the decreasing mass of the rocket as it consumes fuel, the acceleration varies throughout the flight. This variation must be carefully modeled and controlled to ensure the rocket's successful ascent.
Initial Velocity
The initial velocity is the velocity of the rocket at the moment the fuel is ignited. In our textbook problem, this initial velocity is zero as the rocket starts from rest on the launch pad. Throughout the flight, the velocity of the rocket continuously changes due to the constant acceleration imparted by the thrust force.

Understanding the initial velocity is crucial for calculating how the velocity changes over time. With the use of kinematic equations, one can find any future velocity knowing the initial velocity, the time elapsed, and the constant acceleration – under the assumption that the rocket's movement is only influenced by the engine's force and not by any external factors like air resistance or gravity.

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Most popular questions from this chapter

Two crates connected by a rope lie on a horizontal surface (Fig. E5.37). Crate \(A\) has mass \(m_{A}\), and crate \(B\) has mass \(m_{B}\). The coefficient of kinetic friction between each crate and the surface is \(\mu_{\mathrm{k}} .\) The crates are pulled to the right at constant velocity by a horizontal force \(\overrightarrow{\boldsymbol{F}}\). Draw one or more free-body diagrams to calculate the following in terms of \(m_{A}, m_{B},\) and \(\mu_{\mathrm{k}}:\) (a) the magnitude of \(\overrightarrow{\boldsymbol{F}}\) and \((\mathrm{b})\) the tension in the rope connecting the blocks.

A box with mass \(m\) sits at the bottom of a long ramp that is sloped upward at an angle \(\alpha\) above the horizontal. You give the box a quick shove, and after it leaves your hands it is moving up the ramp with an initial speed \(v_{0}\). The box travels a distance \(d\) up the ramp and then slides back down. When it returns to its starting point, the speed of the box is half the speed it started with; it has speed \(v_{0} / 2 .\) What is the coefficient of kinetic friction between the box and the ramp? (Your answer should depend on only \(\alpha\).)

A box of bananas weighing \(40.0 \mathrm{~N}\) rests on a horizontal surface. The coefficient of static friction between the box and the surface is \(0.40,\) and the coefficient of kinetic friction is \(0.20 .\) (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on it? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of \(6.0 \mathrm{~N}\) to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? (e) If the monkey applies a horizontal force of \(18.0 \mathrm{~N},\) what is the magnitude of the friction force and what is the box's acceleration?

The cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of \(100 \mathrm{~m}\). Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every \(60.0 \mathrm{~s}\) ). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs \(882 \mathrm{~N}\) at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

Forces During Chin-ups. When you do a chin-up, you raise your chin just over a bar (the chinning bar), supporting yourself with only your arms. Typically, the body below the arms is raised by about \(30 \mathrm{~cm}\) in a time of \(1.0 \mathrm{~s},\) starting from rest. Assume that the entire body of a \(680 \mathrm{~N}\) person doing chin-ups is raised by \(30 \mathrm{~cm},\) and that half the \(1.0 \mathrm{~s}\) is spent accelerating upward and the other half accelerating downward, uniformly in both cases. Draw a free-body diagram of the person's body, and use it to find the force his arms must exert on him during the accelerating part of the chin-up.

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