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College Algebra : Applications of Quadratic Equations

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Example Question #1 : Applications Of Quadratic Equations

Quadratic equations appear often in physics. The basic kinematic equations for the position of a particle as a function of time , with an initial velocity v_o (a constant) and constant acceleration can be written as,

$x(t)=x_o +v_ot+\frac{1}{2}at^2$

This is a quadratic function in . The function therefore gives the position as a quadratic function of time . If we are dealing with a free-falling object under Earth's gravitational field, we might write this function in the form,

$h(t)=h_o +v_ot +\frac{1}{2}gt^2$

to express the "height" of the object at a given time falling with a constant acceleration g =- 9.8 m/s^2 . Here h_o the initial height (a constant). The units for acceleration are meters-per-square second m/s^2 . The negative acceleration is a convention to signify that the direction of the acceleration is downward.

Find the time required for a ball dropped from a height of 100 m from rest to reach the ground using the quadratic function for height written below,

$h(t)=h_o +v_ot +\frac{1}{2}gt^2$

(Hint, what is the value of the height h(t) when the ball strikes the ground?).

Possible Answers:

$t \approx 4.5 s$

$t \approx 8.3 s$

$t \approx 9.8 s$

$t \approx 20.4 s$

$t \approx 12.1 s$

Correct answer:

$t \approx 4.5 s$

Explanation:

We're given the function,

$h(t)=h_o +v_ot +\frac{1}{2}gt^2$

We know that the gravitational acceleration on earth is:

g =- 9.8 m/s^2 (meters-per-square second)

Because the ball starts at rest, the initial velocity v_o is zero,

v_o =0 (meters-per-second)

We are given the initial height,

h_o = 100 m (meters)

When the ball reaches the ground, the height is "zero," so the value of h(t) is zero at this time, and so we have:

$h(t)=100 + (0)t-\frac{1}{2}(9.8)t^2=0$ (units omitted in the equation).

100-4.9t^2 =0

4.9t^2=100

$t =\sqrt{ \frac{100}{4.9}}\approx 4.5 s$

(Note that although taking the square root of both sides of an equation will produce positive and negative solutions, we ignore the negative solution since "negative time" makes no sense in the context of this problem).

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Example Question #1 : Applications Of Quadratic Equations

A sidewalk on a street corner reaches from a library to a bookstore making an L shape around that corner. The L shaped path has one length that is twice as long as the other. The diagonal path being built between the two buildings will be 102 feet long. How many feet shorter is the diagonal path than the traditional L shaped path around the corner?

Possible Answers:

$12\hspace{1mm} feet$

$22\hspace{1mm}feet$

$35 \hspace{1mm}feet$

$137\hspace{1mm} feet$

The diagonal path would be longer.

Correct answer:

$35 \hspace{1mm}feet$

Explanation:

Use Pythagorean Theorem to solve.

a^2+b^2=c^2

x^2+(2x)^2=(102)^2

$5x^2=10,404\rightarrow x^2=2080.8 \rightarrow x=\pm \sqrt{2080.8}$

$x=\sqrt{2080.8}$

Since the path around the corner is 3x, 3 times the $\sqrt{2080.8}$ is approximately 137. The distance saved is 137-102= 35 feet. Note that we only needed the positive square root since we cannot have a negative distance.

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College Algebra : Applications of Quadratic Equations

Study concepts, example questions & explanations for College Algebra

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Example Questions

Example Question #1 : Applications Of Quadratic Equations

Example Question #1 : Applications Of Quadratic Equations

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