Chapter 1.8, Problem 70PS

Solution

a.

To Determine: The height of the vehicle when it lands on the ramp.

The height of the vehicle when it lands on the ramp is $20ft$ .

Given:

The equation $y=-\frac{1}{640}{x}^2+\frac{1}{4}x+20$

Jumping distance from the ground $=20ft$

Horizontal distance $=x$

Height $=y$

Calculation:

Find the height $r$ of the vehicle when it lands on the ramp.

Consider the diagram,

  

The diagram shows that a stunt motorcycle jumps from one ramp to another ramp 20 feet above the ground.

Jumps between the ramps are modelled by,

  $y=-\frac{1}{640}{x}^2+\frac{1}{4}x+20$

Upon landing on the ramp, the motorcycle's height is $20ft$ , which is the ramp's height:

  $r=20ft$

So, the height of the vehicle when it lands on the ramp is $20ft$ .

b.

To Determine: The distance $d$ between the ramps.

The distance $d$ between the ramps are $160ft$ .

Given:

The equation $y=-\frac{1}{640}{x}^2+\frac{1}{4}x+20$

Jumping distance from the ground $=20ft$

Horizontal distance $=x$

Height $=y$

Calculation:

Find the distance between the ramps $d$ :

Since, from part $\left(a\right)$ its been found that the height $\left(y\right)-axis$ is $20ft$ , substitute $y=20$ in the equation $y=-\frac{1}{640}{x}^2+\frac{1}{4}x+20$ :

  $20=-\frac{1}{640}{x}^2+\frac{1}{4}x+20$

Cancel the similar term:

  $\begin{array}{c}\frac{1}{640}{x}^2=\frac{1}{4}x\\ \frac{x^2}{640}=\frac{x}{4}\\ x^2\cdot 4=640x\end{array}$

Move 4 to the left of $x^2$ :

  $4x^2=640x$

Subtract $640x$ from both sides of the equation:

  $4x^2-640x=0$

Factor $4x$ out of $4x^2-640x$ :

  $4x\left(x-160\right)=0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ :

  $\begin{array}{c}x=0\\ x-160=0\end{array}$

Eliminate the value $x=0$

So, the distance $d$ between the ramps are $160ft$ .

c.

To Determine: The horizontal distance $h$ , the vehicle has travelled when it reaches the maximum height.

The horizontal distance $h$ , the vehicle has travelled when it reaches the maximum height is $80ft$ .

Given:

The equation $y=-\frac{1}{640}{x}^2+\frac{1}{4}x+20$

Jumping distance from the ground $=20ft$

Horizontal distance $=x$

Height $=y$

Calculation:

Find the horizontal distance $h$ , the vehicle has travelled when it reaches the maximum height:

Rewrite the function $y=-\frac{1}{640}{x}^2+\frac{1}{4}x+20$ in vertex form to find $h$ :

The vertex form of a quadratic function is $f\left(x\right)=a\left(x-h\right)2+k$ , where $a,h,$ and $k$ are constants.

Here $h$ represents the horizontal length.

  $\begin{array}{c}y=20-\frac{1}{640}\left(x^2-160x\right)\\ y=20-\frac{1}{640}\left(x^2-160x+{\left(-\frac{160}{2}\right)}^2\right)+\frac{1}{640}{\left(-\frac{160}{2}\right)}^2\\ y=30-\frac{1}{640}{\left(x-80\right)}^2\\ y=-\frac{1}{640}{\left(x-80\right)}^2+30\end{array}$

Since, $h=80$ according to the vertex form, the horizontal distance where the motorcycle has reached maximum height is $80ft$ .

d.

To Determine: The vehicle's maximum height $k$ above the ground.

The vehicle's maximum height $k$ above the ground is $30ft$ .ss

Given:

The equation $y=-\frac{1}{640}{x}^2+\frac{1}{4}x+20$

Jumping distance from the ground $=20ft$

Horizontal distance $=x$

Height $=y$

Calculation:

Find the vehicle's maximum height $k$ above the ground:

The vertex form of a quadratic function is $f\left(x\right)=a\left(x-h\right)2+k$ , where $a,h,$ and $k$ are constants.

Here $k$ represents the maximum height.

From the equation from part $\left(c\right)$ , the equation is:

  $y=-\frac{1}{640}{\left(x-80\right)}^2+30$

Since, $k=30$ according to the vertex form, the maximum height.

above the ground is $30ft$ .