Problem 56 - Algebra 2

Question

In an engineering test, a rocket sled is propelled into a target. The sled's distance \(d\) in meters from the target is given by the formula \(d=-1.5 t^{2}+120,\) where \(t\) is the number of seconds after rocket ignition. How many seconds have passed since rocket ignition when the sled is 10 meters from the target?

Step-by-Step Solution

Answer

Approximately 8.56 seconds.

Step 1: Understand the Given Formula

We are given the formula for the distance from the target as \( d = -1.5t^2 + 120 \), where \( d \) is the distance in meters and \( t \) is the time in seconds. We need to find \( t \) when the sled is 10 meters away from the target.

Step 2: Set Up the Equation

Substitute \( d = 10 \) meters into the formula: \[ 10 = -1.5t^2 + 120 \] This will allow us to solve for \( t \).

Step 3: Rearrange the Equation

To solve for \( t \), first rearrange the equation: \[ -1.5t^2 = 10 - 120 \] This simplifies to: \[ -1.5t^2 = -110 \]

Step 4: Solve for \( t^2 \)

Divide both sides of the equation by \(-1.5\) to isolate \( t^2 \): \[ t^2 = \frac{-110}{-1.5} \] \[ t^2 = 73.33 \]

Step 5: Solve for \( t \)

To find \( t \), take the square root of both sides of the equation: \[ t = \sqrt{73.33} \] \[ t \approx 8.56 \]

Step 6: Interpret the Result

Since \( t = 8.56 \) is a non-negative value, it is a valid solution for the time since time cannot be negative. Thus, approximately 8.56 seconds have passed since rocket ignition when the sled is 10 meters from the target.

Key Concepts

Rocket Sled Distance Problem

The Rocket Sled Distance Problem is an interesting application of quadratic equations in a real-world scenario. Imagine a rocket sled being launched and its distance to a target being tracked over time. The distance is not a constant speed; instead, it forms a non-linear relationship, captured by a quadratic equation.

In our problem, the sled's distance from the target is described by the equation:

• Distance formula: \( d = -1.5t^2 + 120 \)
• Here, \( d \) is the distance in meters, and \( t \) is the time in seconds.

Where this becomes intriguing is how the distance decreases due to the squared term \(-1.5t^2\), representing a deceleration effect in the sled's motion. The rocket sled starts at 120 meters from the target and moves closer over time, ultimately requiring us to find out how long it takes to just barely reach the 10 meter mark.

Solving Quadratic Equations

Solving quadratic equations is a fundamental skill in mathematics. Quadratic equations take the form \( ax^2 + bx + c = 0 \). In the rocket sled problem, we deal with a specific quadratic equation simplified to:

• \( 10 = -1.5t^2 + 120 \)

To find the solution, we need to isolate \( t \):

• Rearrange to: \(-1.5t^2 = -110\)
• Divide by -1.5, getting \(t^2 = 73.33\)

The next step involves finding the square root of 73.33 to determine \( t \). Remember, when solving \( t^2 = 73.33 \), take the positive square root since time cannot be negative. This yields \( t \approx 8.56\), indicating the seconds elapsed.

Time and Distance Problems

Time and distance problems often involve algebraic manipulation to find unknowns, such as time or speed, using given constants or conditions. These problems usually depend on motion equations or distance-time relationships, just as in the case of our rocket sled.

The core of these problems:

• Understanding how variables like time \( t \) and distance \( d \) are related.
• Using equations to represent these relationships.

In our scenario:

• The relationship is quadratic, revealing an accelerating or decelerating pattern of movement.

Grasping the dynamics of these equations helps solve many real-world scenarios, like calculating how long it a speedboat arrives at a dock or a plane reaches a cruising altitude.