QuestionA car traveling at $x$ ft/s takes 1.5 s to react and $\frac{x}{24}$ s to stop. It travels 165 ft total. Find $x$ .

Studdy Solution

STEP 1

Assumptions1. The car is traveling at $x$ feet per second. The driver takes1.5 seconds to react and apply the brake3. After the brake is applied, the car takes $\frac{x}{24}$ seconds to stop4. The average speed of the car after the brake is applied is $\frac{x}{}$ feet per second5. The total distance traveled from the time the driver saw the red light to the time the car comes to a complete stop is165 feet

STEP 2

First, we need to calculate the distance traveled during the reaction time. This can be done by multiplying the speed of the car by the reaction time.
$Distance_{reaction} = Speed \\times Reaction\, time$

STEP 3

Now, plug in the given values for the speed and reaction time to calculate the distance traveled during the reaction time.
$Distance_{reaction} = x \\times1.5$

STEP 4

Calculate the distance traveled during the reaction time.
$Distance_{reaction} =1.x$

STEP 5

Next, we need to calculate the distance traveled after the brake is applied. This can be done by multiplying the average speed of the car after the brake is applied by the time it takes for the car to stop.
$Distance_{braking} = Average\, speed_{braking} \\times Time_{braking}$

STEP 6

Now, plug in the given values for the average speed after braking and the time it takes for the car to stop to calculate the distance traveled after the brake is applied.
$Distance_{braking} = \frac{x}{2} \\times \frac{x}{24}$

STEP 7

Calculate the distance traveled after the brake is applied.
$Distance_{braking} = \frac{x^2}{48}$

STEP 8

Now that we have the distances traveled during the reaction time and after the brake is applied, we can add these together to get the total distance traveled. This should be equal to165 feet.
$Total\, distance = Distance_{reaction} + Distance_{braking} =165$

STEP 9

Plug in the values for the distances traveled during the reaction time and after the brake is applied to get the equation that can be used to find the value of $x$ .
$.5x + \frac{x^2}{48} =165$

STEP 10

Multiply the entire equation by48 to clear the fraction.
$48(.5x) +48(\frac{x^2}{48}) =165 \\times48$

STEP 11

implify the equation to get the final equation.
$72x + x^ =7920$ This corresponds to option D) $x^{}+72 x-7,920$ .

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