Question

A car traveling at 45 miles per hour is brought to a stop, at constant deceleration, 132 feet from where the brakes are applied. (Round your answers to two decimal places.)
Required:
a. How far has the car moved when its speed has been reduced to 30 miles per hour?
b. How far has the car moved when its speed has been reduced to 15 miles per hour?
c. Draw the real number line from 0 to 132, and plot the points found in parts (a) and (b).

Answer 1

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 $v = \sqrt{4356 -33\cdot \Delta s}$, where $v$, in feet per second, and $\Delta s$, in feet, and the graph is included in the image below.

Procedure - Determination of current positions for a car that decelerates uniformly

In this question we must use the following kinematic formula to determine the deceleration rate ($a$), in feet per square second, experimented by the vehicle:

$a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s}$ (1)

Where:

• $\Delta s$ - Travelled distance, in feet.
• $v_{o}$ - Initial speed, in feet per second.
• $v$ - Final speed, in feet per second.

If we know that $\Delta s = 132\,ft$, $v_{o} = 66\,\frac{ft}{s} \left(45\,\frac{mi}{h}\right)$ and $v = 0\,\frac{ft}{s}$, the deceleration rate experimented by the car is:

$a = \frac{\left(0\,\frac{ft}{s}\right)^{2}-\left(66\,\frac{ft}{s} \right)^{2}}{2\cdot (132\,ft)}$

$a = -16.5\,\frac{ft}{s^{2}}$

a) The distance traveled by the car when speed has been reduced to 30 miles per hour

By using (1) and knowing that $v_{o} = 66\,\frac{ft}{s} \,\left(45\,\frac{mi}{h} \right)$, $v = 44\,\frac{ft}{s}\left(30\,\frac{mi}{h} \right)$ and $a = -16.5\,\frac{ft}{s^{2}}$, we find the distance travelled by the vehicle:

$\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$ (1b)

$\Delta s = \frac{\left(44\,\frac{ft}{s} \right)^{2}-\left(66\,\frac{ft}{s} \right)^{2}}{2\cdot \left(-16.5\,\frac{ft}{s^{2}} \right)}$

$\Delta s = 73.333\,ft$

The car moved 73.333 feet when its speed has been reduced to 30 miles per hour. $\blacksquare$

b) The distance traveled by the car when speed has been reduced to 15 miles per hour

By using the same approach used in part (a) and knowing that  $v_{o} = 66\,\frac{ft}{s} \,\left(45\,\frac{mi}{h} \right)$, $v = 22\,\frac{ft}{s}\left(15\,\frac{mi}{h} \right)$ and $a = -16.5\,\frac{ft}{s^{2}}$, we find the distance travelled by the vehicle:

$\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$\Delta s = \frac{\left(22\,\frac{ft}{s} \right)^{2}-\left(66\,\frac{ft}{s} \right)^{2}}{2\cdot \left(-16.5\,\frac{ft}{s^{2}} \right)}$

$\Delta s = 117.333\,ft$

The car moved 117.333 feet when its speed has been reduced to 15 miles per hour. $\blacksquare$

c) Graph of the distance vs speed

In this case we must use (1) in the following form to obtain every speed associated with each travelled distance:

$v = \sqrt{v_{o}^{2}+2\cdot a\cdot \Delta s}$ (1c)

If we know that $v_{o} = 66\,\frac{ft}{s} \,\left(45\,\frac{mi}{h} \right)$ and $a = -16.5\,\frac{ft}{s^{2}}$, then we have the following formula and respective graph.

$v = \sqrt{4356 -33\cdot \Delta s}$ $\blacksquare$

And the graph is presented in the image attached below. $\blacksquare$

To learn more on uniform accelerated motion, we kindly invite to check this verified question: brainly.com/question/12920060