a) The car moved 73.333 feet when its speed has been reduced to 30 miles per hour.
b) The car moved 117.333 feet when its speed has been reduced to 15 miles per hour.
c) The curve is described by [tex]v = \sqrt{4356 -33\cdot \Delta s}[/tex], where [tex]v[/tex], in feet per second, and [tex]\Delta s[/tex], in feet, and the graph is included in the image below.
In this question we must use the following kinematic formula to determine the deceleration rate ([tex]a[/tex]), in feet per square second, experimented by the vehicle:
[tex]a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s}[/tex] (1)
Where:
If we know that [tex]\Delta s = 132\,ft[/tex], [tex]v_{o} = 66\,\frac{ft}{s} \left(45\,\frac{mi}{h}\right)[/tex] and [tex]v = 0\,\frac{ft}{s}[/tex], the deceleration rate experimented by the car is:
[tex]a = \frac{\left(0\,\frac{ft}{s}\right)^{2}-\left(66\,\frac{ft}{s} \right)^{2}}{2\cdot (132\,ft)}[/tex]
[tex]a = -16.5\,\frac{ft}{s^{2}}[/tex]
By using (1) and knowing that [tex]v_{o} = 66\,\frac{ft}{s} \,\left(45\,\frac{mi}{h} \right)[/tex], [tex]v = 44\,\frac{ft}{s}\left(30\,\frac{mi}{h} \right)[/tex] and [tex]a = -16.5\,\frac{ft}{s^{2}}[/tex], we find the distance travelled by the vehicle:
[tex]\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex] (1b)
[tex]\Delta s = \frac{\left(44\,\frac{ft}{s} \right)^{2}-\left(66\,\frac{ft}{s} \right)^{2}}{2\cdot \left(-16.5\,\frac{ft}{s^{2}} \right)}[/tex]
[tex]\Delta s = 73.333\,ft[/tex]
The car moved 73.333 feet when its speed has been reduced to 30 miles per hour. [tex]\blacksquare[/tex]
By using the same approach used in part (a) and knowing that [tex]v_{o} = 66\,\frac{ft}{s} \,\left(45\,\frac{mi}{h} \right)[/tex], [tex]v = 22\,\frac{ft}{s}\left(15\,\frac{mi}{h} \right)[/tex] and [tex]a = -16.5\,\frac{ft}{s^{2}}[/tex], we find the distance travelled by the vehicle:
[tex]\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}[/tex]
[tex]\Delta s = \frac{\left(22\,\frac{ft}{s} \right)^{2}-\left(66\,\frac{ft}{s} \right)^{2}}{2\cdot \left(-16.5\,\frac{ft}{s^{2}} \right)}[/tex]
[tex]\Delta s = 117.333\,ft[/tex]
The car moved 117.333 feet when its speed has been reduced to 15 miles per hour. [tex]\blacksquare[/tex]
In this case we must use (1) in the following form to obtain every speed associated with each travelled distance:
[tex]v = \sqrt{v_{o}^{2}+2\cdot a\cdot \Delta s}[/tex] (1c)
If we know that [tex]v_{o} = 66\,\frac{ft}{s} \,\left(45\,\frac{mi}{h} \right)[/tex] and [tex]a = -16.5\,\frac{ft}{s^{2}}[/tex], then we have the following formula and respective graph.
[tex]v = \sqrt{4356 -33\cdot \Delta s}[/tex] [tex]\blacksquare[/tex]
And the graph is presented in the image attached below. [tex]\blacksquare[/tex]
To learn more on uniform accelerated motion, we kindly invite to check this verified question: https://brainly.com/question/12920060
