91Ó°ÊÓ

A car can be braked to a stop from the autobahn-like speed of \(200 \mathrm{~km} / \mathrm{h}\) in \(170 \mathrm{~m}\). Assuming the acceleration is constant, find its magnitude in (a) SI units and (b) in terms of \(g\). (c) How much time \(T_{b}\) is required for the braking? Your reaction time \(T_{r}\) is the time you require to perceive an emergency, move your foot to the brake, and begin the braking. If \(T_{r}=400 \mathrm{~ms}\), then \((\mathrm{d})\) what is \(T_{b}\) in terms of \(T_{r}\), and (e) is most of the full time required to stop spent in reacting or braking? Dark sunglasses delay the visual signals sent from the eyes to the visual cortex in the brain, increasing \(T_{r} .(\mathrm{f})\) In the extreme case in which \(T_{r}\) is increased by \(100 \mathrm{~ms}\), how much farther does the car travel during your reaction time?

Step by step solution

01

Understanding the Given Parameters

The initial speed of the vehicle, \( v_i = 200 \text{ km/h} \). Convert it to m/s: \( v_i = \frac{200 \times 1000}{3600} = 55.56 \text{ m/s} \). The final speed, \( v_f = 0 \text{ m/s} \). Braking distance, \( d = 170 \text{ m} \). Reaction time, \( T_r = 0.4 \text{ s} \).

02

Calculate Magnitude of Acceleration (SI Units)

Use the kinematic equation \[ v_f^2 = v_i^2 + 2a d \] where \( v_f = 0 \). Solve for acceleration \( a \): \[ 0 = (55.56)^2 + 2a(170) \]\[ a = -\frac{(55.56)^2}{2 \times 170} \approx -9.07 \text{ m/s}^2 \].

03

Calculate Acceleration in Terms of g

Recall that \( g = 9.81 \text{ m/s}^2 \). Express the acceleration \( -9.07 \text{ m/s}^2 \) in terms of \( g \): \[ a_g = \frac{-9.07}{9.81} \approx -0.92g \].

04

Calculate Braking Time \( T_b \)

Use the equation \( v_f = v_i + aT_b \). Solve for \( T_b \):\[ 0 = 55.56 + (-9.07) \cdot T_b \]\[ T_b = \frac{55.56}{9.07} \approx 6.12 \text{ s} \].

05

Braking and Reaction Time Relation

The total time to stop is the sum of braking time and reaction time: \( T_{total} = T_b + T_r \). Since \( T_r = 0.4 \text{ s} \), \( T_{b} = 6.12 \, \text{s} \), so \( T_{total} = 6.12 + 0.4 = 6.52 \text{ s} \).

06

Determine Majority of Total Time

Compare \( T_b \) and \( T_r \): \( T_b = 6.12 \text{ s} \) is the majority of \( T_{total} \), indicating most time is spent braking.

07

Calculate Increased Reaction Time Distance

With new reaction time \( T_{r} = 0.5 \text{ s} \), calculate the distance during reaction:\[ d_{extra} = v_i \cdot (0.5 - 0.4) \]\[ d_{extra} = 55.56 \cdot 0.1 \approx 5.56 \text{ m} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Acceleration

In kinematics, constant acceleration means that the rate of change of velocity remains uniform over time. This implies that every second, the speed changes by a fixed amount, either increasing or decreasing. In the case of a car braking from a high speed, constant negative acceleration, or deceleration, is expected. The deceleration value tells us how quickly the car slows down until it comes to a stop.

To find the magnitude of constant acceleration, kinematic equations like \[ v_f^2 = v_i^2 + 2a d \]can be used, where:

• \( v_f \), the final velocity is usually 0 when the car stops completely.
• \( v_i \) is the initial speed.
• \( a \) is the constant acceleration.
• \( d \) is the distance over which the acceleration occurs.

This way, the magnitude of acceleration can be calculated using information about initial velocity and distance. Remember, knowing the acceleration helps in determining how safe the braking process is.

Braking Distance

Braking distance is the total distance a vehicle travels from the moment the brakes are applied until it comes to a complete stop. It's crucial for road safety, especially at high speeds. Braking distance is a component of the stopping distance, which also includes the distance traveled during the reaction time.

A number of factors can impact braking distance, including:

• Vehicle speed: Higher speeds lead to longer braking distances.
• Road conditions: Wet or icy roads can increase the distance needed.
• Vehicle condition: Worn-out brakes or tires may not stop the car effectively.

Using the information provided, the braking distance in the given exercise was calculated as 170 meters, which helps determine the effectiveness of the braking system and potential need for adjustments at various speeds and conditions.

Reaction Time

Reaction time is the period it takes for a person to respond to a stimulus and begin executing an action. In terms of driving, it is the time between recognizing the need to brake and actually applying the brakes. Everyone's reaction time can be different, and it is influenced by factors like fatigue, distractions, and even the type of sunglasses you wear, as illustrated in the exercise scenario.

Typical reaction times are in the range of 0.2 to 0.9 seconds. During this time, the car continues to travel at its initial speed, adding to the total stopping distance. Calculating additional travel distance during reaction time involves using the formula:\[ d = v imes T_r \]where:

• \( v \) is the velocity.
• \( T_r \) is the reaction time.

Understanding and accounting for reaction time is vital for designing safety features in vehicles, and for drivers to manage safe following distances.

Kinematic Equations

Kinematic equations are mathematical formulas used to predict an object's future position and velocity when the motion involves constant acceleration. These equations allow us to solve various problems related to motion, using known values to find unknown quantities.

The basic kinematic equations in one-dimensional motion under constant acceleration are:

• \( v = v_i + at \)
• \( d = v_i t + \frac{1}{2}a t^2 \)
• \( v_f^2 = v_i^2 + 2ad \)

In the exercise, these equations are crucial for calculating braking time (\( T_b \)) and the acceleration magnitude. Knowing which equation to apply is important for efficiently tackling problems involving motion, and understanding how changes in one quantity affect others in kinematic scenarios.