Question

A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by \(a_y =\) (2.80 m/s\(^3)t\), where the \(+y\)-direction is upward. (a) What is the height of the rocket above the surface of the earth at \(t =\) 10.0 s? (b) What is the speed of the rocket when it is 325 m above the surface of the earth?

Short answer

(a) 466.67 m; (b) 110.68 m/s

Step-by-step solution

Analyzing the problem

We have been given a time-dependent acceleration function, and we need to find the height and speed of the rocket at specific times and heights. The key given function is \( a_y = 2.80t \) m/s³.

Find Velocity Function

To find the velocity function, integrate the acceleration function with respect to time. The acceleration function given is \( a_y = 2.80t \). Integrate: \[ v_y(t) = \int a_y \, dt = \int 2.80t \, dt = 1.40t^2 + C \]Since the rocket starts from rest at \( t = 0 \), \( v_y(0) = 0 \). Therefore, the constant \( C = 0 \). So, \( v_y(t) = 1.40t^2 \).

Find Displacement Function

To find the displacement function, integrate the velocity function with respect to time. \[ y(t) = \int v_y(t) \, dt = \int 1.40t^2 \, dt = \frac{1.40}{3}t^3 + C \]Since the displacement is also 0 at \( t = 0 \), \( y(0) = 0 \). Therefore, the constant \( C = 0 \). Thus, \( y(t) = \frac{1.40}{3}t^3 \).

Calculate Height at t = 10 s

Substitute \( t = 10 \) into the displacement function to find the height of the rocket at 10 seconds:\[ y(10) = \frac{1.40}{3} (10)^3 = \frac{14,000}{3} = 466.67 \, \text{m} \]

Solve for Time when Rocket is 325 m High

Set the displacement equation equal to 325 m and solve for \( t \):\[ \frac{1.40}{3}t^3 = 325 \]Multiply both sides by 3:\[ 1.40t^3 = 975 \]Divide by 1.40:\[ t^3 = \frac{975}{1.40} = 696.43 \]\[ t = \sqrt[3]{696.43} \approx 8.88 \, \text{s} \]

Calculate Speed at 325 m Height

Using the velocity function, find the speed when \( t \approx 8.88 \) s:\[ v_y(8.88) = 1.40(8.88)^2 \]\[ v_y(8.88) \approx 110.68 \, \text{m/s} \]

Key concepts

Rocket Motion

Rocket motion refers to the movement of a rocket as it launches and moves through the atmosphere. This type of motion is studied in physics as part of kinematics, which is the branch of mechanics concerned with the motion of objects. A key component of rocket motion is understanding how forces act on the rocket to change its velocity and direction.

In our exercise, the rocket begins at rest and then accelerates upwards. This initial motion is crucial because it determines how far and how fast the rocket will travel. The forces affecting the rocket include gravity acting downwards and the thrust generated by the rocket's engines pushing it upwards. By calculating these factors over time, we can predict the rocket’s path and determine key aspects like its speed at a certain height or its total displacement after a certain duration.

Time-dependent Acceleration

Time-dependent acceleration is a situation where the acceleration is directly influenced by the passage of time. Unlike constant acceleration, where the acceleration remains the same, time-dependent acceleration changes according to a function of time. This introduces some complexity to solving motion problems because both velocity and displacement are affected by time-varying forces.

In this exercise, the rocket's vertical acceleration, \( a_y = 2.80t \) m/s³, changes linearly with time. This means every second, the rocket's acceleration increases by \( 2.80 \, ext{m/s}^2 \). Understanding this principle allows us to find how the velocity and displacement of the rocket evolve over time. The increasing acceleration indicates that the force exerted by the rocket's engines is also increasing, allowing the rocket to move faster.

Velocity Function

The velocity function represents how an object’s velocity changes over time. In calculus, this is determined by integrating the acceleration function. Since velocity is the integral of acceleration, finding the velocity function involves calculating \[ v_y(t) = \int a_y(t) \, dt. \]

For our rocket problem, this leads to the velocity function \( v_y(t) = 1.40t^2 \). This function shows that the rocket's velocity increases with the square of time, indicating a rapid acceleration at higher values of \( t \). If you know the velocity function, you can easily find the speed at any given time by substituting the time into the function. Since the rocket starts from rest, the velocity was initially zero, and the function was adjusted to include this initial condition. This growing velocity is the result of increasing acceleration over time, propelling the rocket faster with each passing second.

Displacement Function

The displacement function provides information about an object's position over time. It is derived by integrating the velocity function, which reveals how the object's position changes as it moves. Mathematically, this is determined by: \[ y(t) = \int v_y(t) \, dt. \]

For the problem in question, the displacement function is \( y(t) = \frac{1.40}{3}t^3 \). This implies that the height or position of the rocket above the Earth's surface is related to the cube of time. The cubic relationship highlights the effect of the increasing velocity and acceleration on the rocket's height over time.

Using this function, we calculated the height of the rocket at a specific time or found the time when the rocket reaches a certain height. Understanding displacement is critical because it tells us exactly where the rocket is in its journey at any given second, allowing us to predict its future position.