Problem 51
Question
A light-rail train traveling at \(\frac{14.67 \mathrm{ft}}{1 \mathrm{~s}}\) is accelerating at a constant rate of \(\frac{1.27 \mathrm{ft}}{1 \mathrm{~s}^{2}}\). Find the time for the train to travel \(5280 \mathrm{ft}\) by solving the equation \(5280 \mathrm{ft}=\left(\frac{14.67 \mathrm{ft}}{1 \mathrm{~s}}\right) t+\frac{1}{2}\left(\frac{1.27 \mathrm{ft}}{1 \mathrm{~s}^{2}}\right) t^{2}\). Round to the nearest whole number.
Step-by-Step Solution
Verified Answer
The train will take approximately 80 seconds.
1Step 1: Write down the given equation
The equation provided is: \[ 5280 = 14.67 t + \frac{1}{2} (1.27) t^2 \]
2Step 2: Simplify the equation
We can combine the terms to simplify the equation: \[ 5280 = 14.67 t + 0.635 t^2 \]
3Step 3: Rearrange the equation
Move all terms to one side of the equation to set it to zero: \[ 0.635 t^2 + 14.67 t - 5280 = 0 \]
4Step 4: Use the quadratic formula
To solve for t, use the quadratic formula \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where a=0.635, b=14.67, and c=-5280.
5Step 5: Calculate the discriminant
First, calculate the discriminant \[ b^2 - 4ac \] \[ 14.67^2 - 4 \times 0.635 \times (-5280) \] \[ 215.0289 + 13408.8 = 13623.8289 \]
6Step 6: Compute the roots
Now compute the roots of the quadratic equation using the discriminant: \[ t = \frac{-14.67 \pm \sqrt{13623.8289}}{2 \times 0.635} \] \[ t = \frac{-14.67 \pm 116.7045}{1.27} \]
7Step 7: Find the positive root
Calculate both solutions and choose the positive value for the time: \[ t = \frac{-14.67 + 116.7045}{1.27} \approx 80 \]
Key Concepts
algebraquadratic formuladistance, rate, and time
algebra
Algebra is a branch of mathematics that uses symbols to represent numbers and quantities. These symbols are often letters called variables. In solving quadratic equations, we manipulate algebraic expressions to find the value of the unknown variable (in this case, time, or t). Quadratic equations have the form:
\[ ax^2 + bx + c = 0 \]
Where:
Variables and constants can be combined using operations like addition, subtraction, multiplication, and division to form equations. By rearranging equations and combining like terms, we can solve for the variables. In our problem, we had to simplify and rearrange the terms to solve for t.
\[ ax^2 + bx + c = 0 \]
Where:
- a, b, and c are constants
- x is the variable being solved for
Variables and constants can be combined using operations like addition, subtraction, multiplication, and division to form equations. By rearranging equations and combining like terms, we can solve for the variables. In our problem, we had to simplify and rearrange the terms to solve for t.
quadratic formula
The quadratic formula is a standard method used to solve quadratic equations. It is especially useful when factoring is complex or impossible. The quadratic formula is:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here's how it applies to our problem:
The discriminant, \( b^2 - 4ac \), determines the nature of the roots (solutions). If the discriminant is positive, there will be two real roots. We calculated:
\[t = \frac{-14.67 \pm \sqrt{13623.8289}}{1.27} \]
Finally, we selected the positive root as the meaningful solution for time.
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here's how it applies to our problem:
- a = 0.635
- b = 14.67
- c = -5280
The discriminant, \( b^2 - 4ac \), determines the nature of the roots (solutions). If the discriminant is positive, there will be two real roots. We calculated:
\[t = \frac{-14.67 \pm \sqrt{13623.8289}}{1.27} \]
Finally, we selected the positive root as the meaningful solution for time.
distance, rate, and time
In mathematics and physics, the relationship between distance, rate (speed), and time is fundamental. The basic formula connecting these quantities is:
\[ \text{Distance} = \text{Rate} \times \text{Time} \]
For our train problem, the distance traveled by the train is given as 5280 ft. We start with an initial rate of 14.67 ft/s and an acceleration of 1.27 ft/s^2. Using these in the quadratic equation:
\[ 5280 = 14.67 t + \frac{1}{2}(1.27) t^2 \]
gives us a formula that accounts for both the initial speed and the accelerated movement over time. This ensures that we calculate the exact time needed for the train to cover the asked distance.
\[ \text{Distance} = \text{Rate} \times \text{Time} \]
For our train problem, the distance traveled by the train is given as 5280 ft. We start with an initial rate of 14.67 ft/s and an acceleration of 1.27 ft/s^2. Using these in the quadratic equation:
\[ 5280 = 14.67 t + \frac{1}{2}(1.27) t^2 \]
gives us a formula that accounts for both the initial speed and the accelerated movement over time. This ensures that we calculate the exact time needed for the train to cover the asked distance.