Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

On the surface of the moon, the acceleration of gravity is \(-5.28\) feet per second per second. If an object is thrown upward from an initial height of 1000 feet with a velocity of 56 feet per second, find its velocity and height \(4.5\) seconds later.

Step-by-Step Solution

Answer

Velocity is 32.24 feet/second; Height is 1198.54 feet.

Step 1: Understanding the Problem

We need to find the velocity and height of an object 4.5 seconds after it is thrown upwards. The initial height is 1000 feet, and the initial velocity is 56 feet/second. The acceleration due to gravity on the moon is \(-5.28\) feet/second².

Step 2: Write the Velocity Equation

The velocity of an object under constant acceleration is given by the formula: \[ v(t) = v_0 + at \] where \(v_0\) is the initial velocity and \(a\) is the acceleration. Here, \(v_0 = 56\) feet/second and \(a = -5.28\) feet/second².

Step 3: Calculate the Velocity at 4.5 Seconds

Plug in the values into the velocity equation:\[ v(4.5) = 56 + (-5.28) \times 4.5 \]Calculate the result: \[ v(4.5) = 56 - 23.76 = 32.24 \text{ feet/second} \]

Step 4: Write the Height Equation

The height of an object under constant acceleration is given by: \[ h(t) = h_0 + v_0t + \frac{1}{2}at^2 \] where \(h_0\) is the initial height. Here, \(h_0 = 1000\) feet, \(v_0 = 56\) feet/second, and \(a = -5.28\) feet/second².

Step 5: Calculate the Height at 4.5 Seconds

Substitute the values into the height equation:\[ h(4.5) = 1000 + 56 \times 4.5 + \frac{1}{2} \times (-5.28) \times (4.5)^2 \] Calculate step by step: - \(56 \times 4.5 = 252\) - \((4.5)^2 = 20.25\) - \(\frac{1}{2} \times (-5.28) \times 20.25 = -53.46\) - Combine these: \[ h(4.5) = 1000 + 252 - 53.46 = 1198.54 \text{ feet} \]

Key Concepts

Acceleration Due to Gravity

The acceleration due to gravity is a constant force that affects all objects on celestial bodies like the moon, Earth, Mars, and so on. On the moon, this acceleration is significantly lower than on Earth because the moon has less mass and therefore a weaker gravitational pull. In the given exercise, the acceleration due to gravity on the moon is

• a = -5.28 feet/second² **

This negative sign indicates that gravity acts in the opposite direction of an object that is moving upwards. This means that when an object is thrown upwards, gravity will slow it down. It also means that when the object eventually falls back, gravity will speed it up.
Understanding the specific value of gravitational acceleration on the celestial body in question, like the moon in this case, is crucial for accurately predicting motion parameters like speed and position over time.

Velocity Equation

The velocity equation is a fundamental relation in kinematics used to find the speed of an object at any given time when its initial velocity and the acceleration acting on it are known. The formula is:\[v(t) = v_0 + at\]In simple words, this equation helps you find the velocity at any time by adding the initial velocity to the product of acceleration and time.

• **\(v_0\)** is the initial velocity, which is given as 56 feet/second here.
• **\(a\)** is the acceleration due to gravity,
• **\(t\)** represents time, which is 4.5 seconds in this scenario.

This equation considers the initial speed and then adjusts for the constant acceleration experienced over time. Once understood, it's a powerful tool for solving various physics problems involving motion.

Height Equation

The height equation in kinematics provides a way to determine how high or low an object is at a particular point in time when moving vertically. This is crucial when an object is subject to gravity, like in our exercise on the moon. The height equation is:\[h(t) = h_0 + v_0 t + \frac{1}{2} at^2\]This equation blends initial height, initial velocity, and the effect of gravitational acceleration over time.

• **\(h_0\)** is the initial height, 1000 feet in our exercise.
• **\(v_0\)** is the starting velocity, 56 feet per second for us.
• **\(a\)** is the acceleration, negative since it acts downward.
• **\(t\)** is the time passed, 4.5 seconds here.

The equation first accounts for where the object started, adds the effect of its initial speed over the given time, and finally adjusts this with the impact of gravity, which can either increase or decrease the object's position depending on the direction of motion.

Initial Conditions

Initial conditions set the stage for analyzing motion, providing a snapshot that defines where and how fast an object starts moving. For any kinematic problem, knowing the initial conditions ensures accurate prediction of future motion.

• **Initial Velocity (\(v_0\))**: In our problem, the object starts with an initial velocity of 56 feet per second upwards.
• **Initial Height (\(h_0\))**: Here, the object is launched from a height of 1000 feet.

Initial conditions directly feed into kinematic equations for velocity and height, influencing the calculated results of where an object might be or how fast it's moving at any given future point. While solving problems, always identify these values first to guide the use of formulas accurately.