Problem 39 - University Physics with Modern Physics

Question

A rocket is fired at an angle from the top of a tower of height \(h_0\) = 50.0 m. Because of the design of the engines, its position coordinates are of the form \(x(t) = A + Bt^2 \)and \(y(t) = C + Dt^3\), where \(A, B, C,\) and \(D\) are constants. The acceleration of the rocket 1.00 s after firing is \(\vec{a} = (4.00 \hat{i}+ 3.00\hat{j}) m/s^2\). 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

Answer

Constants: \(A = 0\), \(B = 2\), \(C = 50\), \(D = 0.5\). Velocity at 10s: \(v=151.3\) m/s. Position at 10s: \((200, 550)\) m.

Step 1: Analyze Given Information

From the problem, we know:- Height of the tower, \(h_0 = 50.0 \text{ m}\).- Position functions: \(x(t) = A + Bt^2\) and \(y(t) = C + Dt^3\).- Acceleration at \(t = 1 \text{ s}\): \(\vec{a} = (4.00 \hat{i} + 3.00 \hat{j})\, \text{m/s}^2\).- Origin is at the base of the tower. From this information, identify the need to find the constants \(A, B, C,\) and \(D\), the velocity and acceleration immediately after firing, velocity after 10 s, and the position after 10 s.

Step 2: Determine Acceleration Expressions

Acceleration is the second derivative of position with respect to time:- From \(x(t) = A + Bt^2\), the velocity is \(v_x(t) = \frac{dx}{dt} = 2Bt\), and acceleration \(a_x(t) = \frac{d^2x}{dt^2} = 2B\).- From \(y(t) = C + Dt^3\), the velocity is \(v_y(t) = \frac{dy}{dt} = 3Dt^2\), and acceleration \(a_y(t) = \frac{d^2y}{dt^2} = 6Dt\).

Step 3: Solve for Constants B and D

Given \(a_x(1) = 4 \text{ m/s}^2\) and \(a_y(1) = 3 \text{ m/s}^2\):- For x-component: \(2B = 4\), thus, \(B = 2 \text{ m/s}^2\).- For y-component: \(6D(1) = 3\), thus, \(D = 0.5 \text{ m/s}^3\).

Step 4: Solve for Constants A and C

At \(t=0\), the position is at the top of the tower:- \(x(0) = A = 0\) because the origin is directly below.- \(y(0) = C = h_0 = 50.0 \text{ m}\).

Step 5: Initial Acceleration and Velocity

At \(t = 0\):- \(a_x(0) = 2B = 0 \text{ m/s}^2\) and \(a_y(0) = 0 \text{ m/s}^2\).- \(v_x(0) = 2B(0) = 0 \text{ m/s}\) and \(v_y(0) = 3D(0)^2 = 0 \text{ m/s}\). Thus, both initial acceleration and velocity are zero.

Step 6: Velocity at t = 10 s

Calculate the components:- \(v_x(10) = 2B(10) = 20 \text{ m/s}\).- \(v_y(10) = 3D(10)^2 = 3 \times 0.5 \times 100 = 150 \text{ m/s}\).- Total speed: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 150^2} = 151.3 \text{ m/s}\).

Step 7: Position at t = 10 s

Substitute \(t = 10\) into the position equations:- \(x(10) = A + Bt^2 = 0 + 2 \times 100 = 200 \text{ m}\).- \(y(10) = C + Dt^3 = 50 + 0.5 \times 1000 = 550 \text{ m}\).- Position vector: \(\vec{r}(10) = (200 \hat{i} + 550 \hat{j}) \text{ m}\).

Key Concepts

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces causing the motion. It focuses on describing the motion in terms of displacement, velocity, and acceleration, often through equations derived from these quantities. In the case of the rocket fired from the top of a tower, kinematic equations help determine its trajectory and speed over time. When working with such equations, time-dependent position functions given, as in this example, allow us to calculate various quantities by differentiation or substitution.
This includes determining initial and final velocities, accelerations, as well as distances traveled (positions). All these calculations form the foundation of analyzing projectile motion in a kinematic scenario.

Acceleration

Acceleration is the rate at which an object changes its velocity. In the context of projectile motion, acceleration can be uniform or vary with time. In this exercise, the acceleration of the rocket is represented by the vector \( \vec{a} = (4.00 \hat{i} + 3.00 \hat{j}) \text{ m/s}^2 \). The components indicate change in velocity in both the x-direction and y-direction per second. Understanding acceleration involves recognizing it as second derivative of position regarding time:

• For x-direction: \( a_x(t) = \frac{d^2x}{dt^2} = 2B \)
• For y-direction: \( a_y(t) = \frac{d^2y}{dt^2} = 6Dt \)

Acceleration in time-dependent equations allows us to derive constants (like B and D here) crucial for comprehending overall motion and helps predict velocity changes over time.

Initial Conditions

Every problem in kinematics starts by establishing initial conditions, which are the values at the start of observation, typically at time t = 0. These include the initial position, velocity, and sometimes acceleration. In this exercise, the rocket begins at the top of a tower 50 meters above the ground (\( y(0) = 50 \text{ m} \)).

• It was deduced that its position aligns directly above the base, hence \( x(0) = 0 \).
• The initial velocities in both x and y directions were calculated as zero due to non-zero velocity terms developing over time through their respective equations.

Initial conditions set the stage for solving the complete motion, allowing one to plug values into kinematic equations to find desired quantities over time.

Velocity Components

Breaking down motion into velocity components helps in understanding the vector nature of velocity in two-dimensional motion like projectile motion. This includes separate treatment of horizontal (x-component) and vertical (y-component) velocities.
The rocket's velocity components can be determined from the position functions by finding their first derivatives:

• \( v_x(t) = \frac{dx}{dt} = 2Bt \)
• \( v_y(t) = \frac{dy}{dt} = 3Dt^2 \)

At t = 10 seconds, substitution into these equations yields components \( v_x = 20 \text{ m/s} \) and \( v_y = 150 \text{ m/s} \). These components can be combined using the Pythagorean theorem to find the rocket's total velocity magnitude: \( v = \sqrt{v_x^2 + v_y^2} = 151.3 \text{ m/s} \). Understanding individual components offers insights into how an object moves in each direction.