Question

the speed s in miles per hour that a car is traveling when it goes into a skid can be estimated by using the formula ( s = sqrt{30fd} ), where ( f ) is the coefficient of friction and ( d ) is the length of the skid marks in feet. after an accident, a driver claims to have been traveling the speed limit of 55 miles per hour. the coefficient of friction under accident conditions was 0.8. is the driver telling the truth about his speed if the actual skid marks is 125 feet? explain.
drag the expressions or words into the boxes to explain.
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____ telling the truth about his speed. if the driver had been traveling at 55 miles per hour, the skid marks would measure about __ feet. because the skid marks actually measure 125 feet, the driver was probably driving close to ____ miles per hour.
options: 55, is not, is, 60, 131

Explanation:

Step1: Substitute values into formula

We know the formula for speed \( s = \sqrt{30fd} \), where \( f = 0.8 \) (coefficient of friction) and \( d = 125 \) (length of skid marks in feet). Substitute these values into the formula:
\( s=\sqrt{30\times0.8\times125} \)

Step2: Calculate inside the square root

First, calculate \( 30\times0.8\times125 \). We can rearrange the multiplication for easier calculation: \( 30\times(0.8\times125) \). Since \( 0.8\times125 = 100 \), then \( 30\times100 = 3000 \). So now we have \( s=\sqrt{3000} \)

Step3: Approximate the square root

We know that \( \sqrt{2916}=54 \) and \( \sqrt{3025}=55 \). Since \( 3000 \) is between \( 2916 \) and \( 3025 \), \( \sqrt{3000}\approx54.77 \), which is close to \( 55 \) (the speed limit the driver claims). Wait, but let's check the calculation again. Wait, \( 30\times0.8 = 24 \), then \( 24\times125 = 3000 \). \( \sqrt{3000}\approx54.77 \), which is approximately \( 55 \). But wait, maybe I made a mistake. Wait, the formula is \( s=\sqrt{30fd} \), \( f = 0.8 \), \( d = 125 \). So \( 30\times0.8 = 24 \), \( 24\times125 = 3000 \). \( \sqrt{3000}\approx54.77 \), which is about \( 55 \). So if the driver claims to be going 55, and the calculated speed is about 55, then the skid marks would be about 125 feet (since when we plug in \( s = 55 \), \( 55=\sqrt{30\times0.8\times d} \), square both sides: \( 55^2 = 24d \), \( 3025 = 24d \), \( d=\frac{3025}{24}\approx126.04 \). Oh! Wait, I had the substitution reversed. Wait, the problem is: the driver claims speed \( s = 55 \), and actual skid marks \( d = 125 \). Let's check what speed corresponds to \( d = 125 \) and \( f = 0.8 \). So \( s=\sqrt{30\times0.8\times125}=\sqrt{3000}\approx54.77\approx55 \). But when we calculate what \( d \) should be for \( s = 55 \): \( 55=\sqrt{30\times0.8\times d} \), square both sides: \( 3025 = 24d \), \( d=\frac{3025}{24}\approx126.04 \). So the actual skid marks are 125, which is close to 126.04. So the driver's claimed speed of 55 is close to the calculated speed from \( d = 125 \) (which is ~54.77, close to 55). So the driver is telling the truth? Wait, the drag and drop options: "The driver [is not] telling the truth. If the driver had been traveling at 55 miles per hour, the skid marks would measure about [126] feet. Because the skid marks actually measure 125 feet, the driver was probably driving close to [55] miles per hour." Wait, let's recalculate the expected skid marks for \( s = 55 \):

\( s = 55 \), \( f = 0.8 \)

From \( s=\sqrt{30fd} \), solve for \( d \):

\( s^2 = 30fd \)

\( d=\frac{s^2}{30f} \)

Substitute \( s = 55 \), \( f = 0.8 \):

\( d=\frac{55^2}{30\times0.8}=\frac{3025}{24}\approx126.04 \approx 126 \) feet.

So if the actual skid marks are 125 feet, which is close to 126 feet, then the driver's speed is close to 55. So the driver is not (wait, no: if the expected skid marks for 55 is 126, and actual is 125, which is close, so the driver's speed is close to 55. So the driver is telling the truth? Wait, the options are: "The driver [is not] telling the truth. If the driver had been traveling at 55 miles per hour, the skid marks would measure about [126] feet. Because the skid marks actually measure 125 feet, the driver was probably driving close to [55] miles per hour." Wait, maybe the correct drag and drop is:

The driver [is] telling the truth? No, wait, when we calculate \( d \) for \( s = 55 \), we get \( d\approx126 \), and actual \( d = 125 \), which is very close. So the driver's claimed speed of 55 is close to the speed calculated from \( d = 125 \) (which is \( s=\sqrt{30\times0.8\times125}\approx54.77\approx55 \)). So the driver is telling the truth, and the skid marks would be about 126 feet if going 55, and actual is 125, so close to 55.

Wait, let's do the calculation for \( s \) when \( d = 125 \) and \( f = 0.8 \):

\( s=\sqrt{30\times0.8\times125}=\sqrt{3000}\approx54.77\approx55 \)

So the speed is approximately 55, which matches the driver's claim. So the driver is telling the truth. But the drag and drop options: the first blank: "is not" or "is"? Wait, maybe I messed up. Wait, when \( s = 55 \), \( d=\frac{55^2}{30\times0.8}=\frac{3025}{24}\approx126 \). Actual \( d = 125 \), which is 1 less than 126. So the calculated speed for \( d = 125 \) is \( \sqrt{30\times0.8\times125}=\sqrt{3000}\approx54.77\approx55 \). So the driver's claimed speed is 55, and the actual speed is about 55, so the driver is telling the truth. The skid marks for 55 would be about 126 feet, and actual is 125, which is close, so the driver was probably driving close to 55.

So the drag and drop:

The driver [is] telling the truth. If the driver had been traveling at 55 miles per hour, the skid marks would measure about [126] feet. Because the skid marks actually measure 125 feet, the driver was probably driving close to [55] miles per hour.

Wait, but the options given are: 126, 55, is not, is, 60, 131.

So let's fill in:

The driver [is] telling the truth. If the driver had been traveling at 55 miles per hour, the skid marks would measure about [126] feet. Because the skid marks actually measure 125 feet, the driver was probably driving close to [55] miles per hour.

Wait, but maybe the first blank is "is not"? No, because the calculated speed from \( d = 125 \) is ~55, which is the driver's claim. So the driver is telling the truth.

Answer:

The driver \(\boldsymbol{\text{is}}\) telling the truth. If the driver had been traveling at 55 miles per hour, the skid marks would measure about \(\boldsymbol{126}\) feet. Because the skid marks actually measure 125 feet, the driver was probably driving close to \(\boldsymbol{55}\) miles per hour.