Math Problem Statement

  1. a car is traveling at 120km/h (75mph). when applied, the braking system can stop the car with an acceleration of 9m/s^2. the typical reaction time for an alert driver is 0.5s versus 2s for a sleepy driver. assuming a typical car length of 5m, calculate the number of additional car lengths it takes the sleepy driver to stop compared to the alert driver.

Solution

To solve this problem, we need to calculate the stopping distance for both the alert and sleepy drivers. The stopping distance is composed of two parts:

  1. Reaction distance: The distance traveled during the driver's reaction time before the brakes are applied.
  2. Braking distance: The distance the car travels after the brakes are applied until it comes to a complete stop.

Step 1: Convert Speed to Meters per Second

The car's speed is given as 120 km/h. First, we convert this to meters per second: Speed=120×10003600=33.33 m/s\text{Speed} = 120 \times \frac{1000}{3600} = 33.33 \text{ m/s}

Step 2: Calculate the Reaction Distance

The reaction distance is calculated by multiplying the speed of the car by the reaction time.

For the alert driver (reaction time = 0.5 s): Reaction Distancealert=33.33×0.5=16.67 m\text{Reaction Distance}_{\text{alert}} = 33.33 \times 0.5 = 16.67 \text{ m}

For the sleepy driver (reaction time = 2 s): Reaction Distancesleepy=33.33×2=66.67 m\text{Reaction Distance}_{\text{sleepy}} = 33.33 \times 2 = 66.67 \text{ m}

Step 3: Calculate the Braking Distance

The braking distance is calculated using the formula: Braking Distance=v22a\text{Braking Distance} = \frac{v^2}{2a} where v=33.33v = 33.33 m/s and a=9a = 9 m/s².

Braking Distance=(33.33)22×9=1110.891861.71 m\text{Braking Distance} = \frac{(33.33)^2}{2 \times 9} = \frac{1110.89}{18} \approx 61.71 \text{ m}

Step 4: Calculate the Total Stopping Distance

The total stopping distance is the sum of the reaction distance and the braking distance.

For the alert driver: Total Stopping Distancealert=16.67+61.71=78.38 m\text{Total Stopping Distance}_{\text{alert}} = 16.67 + 61.71 = 78.38 \text{ m}

For the sleepy driver: Total Stopping Distancesleepy=66.67+61.71=128.38 m\text{Total Stopping Distance}_{\text{sleepy}} = 66.67 + 61.71 = 128.38 \text{ m}

Step 5: Calculate the Additional Car Lengths

The additional distance the sleepy driver travels compared to the alert driver is: Additional Distance=128.3878.38=50 m\text{Additional Distance} = 128.38 - 78.38 = 50 \text{ m}

Given that the typical car length is 5 meters, the number of additional car lengths is: Additional Car Lengths=505=10 car lengths\text{Additional Car Lengths} = \frac{50}{5} = 10 \text{ car lengths}

Final Answer

The sleepy driver takes 10 additional car lengths to stop compared to the alert driver.


Would you like more details or have any questions?

Here are some related questions to consider:

  1. What factors influence braking distance other than speed and reaction time?
  2. How would the stopping distance change if the car's speed were reduced by half?
  3. How does the coefficient of friction affect the braking distance?
  4. What safety features can reduce the stopping distance in modern vehicles?
  5. How would the stopping distance change if the car were on a wet road?

Tip: Always maintain a safe following distance when driving, as reaction time and stopping distance are critical for preventing accidents.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Physics of Motion
Reaction Time

Formulas

Braking Distance Formula: d = v^2 / (2a)

Theorems

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Suitable Grade Level

High School