91Ó°ÊÓ

A rocket is fired at an angle from the top of a tower of height \(h_{0}=50.0 \mathrm{m} .\) Because of the design of the engines, its position coordinates are of the form \(x(t)=A+B t^{2}\) and \(y(t)=C+D t^{3},\) where \(A, B, C,\) and \(D\) are constants. Furthermore, the acceleration of the rocket 1.00 s after firing is \(\vec{a}=(4.00 \hat{\imath}+3.00 \hat{J}) \mathrm{m} / \mathrm{s}^{2} .\) Take the origin of coordinates to be at the base of the tower. (a) Find the constants \(A, B, C,\) and \(D,\) including their SI units. (b) At the instant after the rocket is fired, what are its acceleration vector and its velocity? (c) What are the \(x\) -and \(y\) -components of the rocket's velocity 10.0 s after it is fired, and how fast is it moving? (d) What is the position vector of the rocket 10.0 s after it is fired?

Step by step solution

01

Analyze the Coordinate Functions

The position functions given are \( x(t) = A + B t^2 \) and \( y(t) = C + D t^3 \). These functions suggest that acceleration will be the derivative of velocity, which is the derivative of position with respect to time.

02

Derive Velocity Functions

The velocity functions are the first derivatives of the position functions: \( v_x(t) = \frac{d}{dt}x(t) = 2Bt \) and \( v_y(t) = \frac{d}{dt}y(t) = 3Dt^2 \).

03

Derive Acceleration Functions

The acceleration functions are the derivatives of the velocity functions: \( a_x(t) = \frac{d}{dt}v_x(t) = 2B \) and \( a_y(t) = \frac{d}{dt}v_y(t) = 6Dt \). At \( t = 1 \text{ s} \), \( a_x(1) = 4.00 \text{ m/s}^2 \) and \( a_y(1) = 3.00 \text{ m/s}^2 \).

04

Solve for Constants B and D

From the equations \( a_x(1) = 2B = 4.00 \text{ m/s}^2 \) and \( a_y(1) = 6D \cdot 1 = 3.00 \text{ m/s}^2 \), solving gives \( B = 2.00 \text{ m/s}^2 \) and \( D = 0.5 \text{ m/s}^3 \).

05

Determine Constants A and C

Since the rocket is at the top of the tower when \( t = 0 \), \( x(0) = A = 0 \), as the origin is at the base. Similarly, \( y(0) = C = 50.0 \text{ m} \).

06

Acceleration at Firing

At the instant after firing, \( t = 0 \) and thus \( a_x(0) = 2B = 4.00 \text{ m/s}^2 \) and \( a_y(0) = 0 = 0 \text{ m/s}^2 \). The acceleration vector is \((4.00 \hat{i} + 0 \hat{j})\text{ m/s}^2.\)

07

Velocity at Firing

At \( t = 0 \), \( v_x(0) = 0 \) and \( v_y(0) = 0 \). The velocity vector is \((0 \hat{i} + 0 \hat{j})\text{ m/s}.\)

08

Velocity at t = 10 seconds

At \( t = 10 \text{ s} \), \( v_x(10) = 2B \cdot 10 = 40.0 \text{ m/s} \) and \( v_y(10) = 3D \cdot 10^2 = 150.0 \text{ m/s} \). The magnitude of velocity is \( \sqrt{40^2 + 150^2} = 155.2 \text{ m/s}.\)

09

Position at t = 10 seconds

At \( t = 10 \text{ s} \), \( x(10) = A + B \cdot 10^2 = 200.0 \text{ m} \) and \( y(10) = C + D \cdot 10^3 = 550.0 \text{ m} \). Position vector is \((200 \hat{i} + 550 \hat{j})\text{ m}.\)

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Key Concepts

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

Coordinate Systems

In physics, coordinate systems are fundamental when analyzing movements such as projectile motion. For this problem, the origin of the coordinate system is set at the base of the tower. This means all measurements are relative to this point. The rocket's motion is described in terms of its position along two axes: the horizontal axis (x-axis) and the vertical axis (y-axis). This setup helps us break down the problem into manageable parts and relate them to real-world distances.

Using this coordinate system, the position of the rocket can be precisely determined at any given time using the functions provided, allowing us to understand its trajectory. Also, by defining the origin at the base of the tower, we simplify the math involved in finding height and distance changes.

• X-axis: Horizontal movement.
• Y-axis: Vertical movement, from the base of the tower upwards.

Acceleration

Acceleration is a crucial concept in projectile motion, referring to how quickly an object's velocity changes. Here, the rocket's acceleration at a specific time (1 second after firing) is known: \( \vec{a} = (4.00 \hat{\imath} + 3.00 \hat{\jmath}) \mathrm{m/s}^2 \). This vector tells us how rapidly the rocket's speed is changing in both the horizontal and vertical directions.

Knowing the acceleration vector helps determine the constants in the position equations, as acceleration is derived from the velocity functions, which in turn are derived from the position functions. Specifically, from the equations given, we calculate the constants B and D, which describe the time-dependent change of acceleration.

• Horizontal Acceleration (\(a_x\)): Shows how horizontal speed increases. Derived from \(v_x(t)\).
• Vertical Acceleration (\(a_y\)): Shows how vertical speed increases. Derived from \(v_y(t)\).

Velocity Calculations

Velocity describes the speed and direction of the rocket's movement and is crucial for determining where the rocket is at any given moment. It is derived from the position functions: \(v_x(t) = 2Bt\) and \(v_y(t) = 3Dt^2 \). Velocity indicates how the rocket changes its position over time in both horizontal and vertical components.

At time \(t = 0\), right after the rocket is fired, both velocity components are zero, meaning the rocket starts from rest. As time progresses, velocity increases depending on the constants B and D, which were found using information about the acceleration. At 10 seconds, the velocity components are calculated to understand how fast the rocket is moving in each direction, and its overall speed.

• Velocity at \(t=10 \text{s}\): Key to evaluating the rocket's speed at this specific moment.
• Overall Speed: Calculated using the Pythagorean theorem to combine both components.

Position Vector

The position vector of the rocket provides a complete description of its location in the coordinate system, being dependent on the time function. It is determined by evaluating the position functions \(x(t) = A + Bt^2\) and \(y(t) = C + Dt^3\) at a given time. These expressions give the rocket's position in terms of horizontal and vertical distances from the origin.

At \(t=10 \text{s}\), the calculated values are substituted into these equations to find the position vector of the rocket. This vector gives the definitive position along x and y axes, indicating where the rocket is within our defined space.

• Horizontal Position (\(x\)): Indicates how far from the tower the rocket has traveled.
• Vertical Position (\(y\)): Shows the rocket's height above the base.
• Position Vector: Represents the \( (x\hat{i} + y\hat{j})\) in meters.