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Chapter 2: Problem 82

A car is driving at a speed of \(40 \mathrm{~km} / \mathrm{h}\) toward an intersection just as the light changes from green to yellow. If the driver has a reaction time of \(0.75 \mathrm{~s}\) and the braking acceleration of the car is \(-5.5 \mathrm{~m} / \mathrm{s}^{2}\), find the minimum distance \(x_{\min }\) the car travels after the light changes before coming to a stop.

Short Answer

Expert verified
The minimum stopping distance for the car after the light changes is \(19.6 \mathrm{~m}\).

Step by step solution

01

Convert Speed from km/h to m/s

Note the speed of the car (40 km/h) needs to be converted into m/s by multiplying it with \( \frac{1000}{3600} \). Therefore, the speed of the car in m/s is \( 40 \times \frac{1000}{3600} = 11.11 \mathrm{~m/s} \)
02

Calculate Distance Covered During Reaction Time

During the driver's reaction time, the car moves with a uniform speed. So, using the equation of motion \( s = ut + \frac{1}{2}at^2 \), where \(s\) is the distance travelled, \(u\) is the initial speed, \(a\) is acceleration and \(t\) is time, we find that \(s_1 = u \times t \) as acceleration is zero during this period. Thus, the distance covered during the driver's reaction time is \(s_1 = 11.11 \mathrm{~m/s} \times 0.75 \mathrm{~s} = 8.33 \mathrm{~m} \).
03

Calculate Distance Covered During Braking

Under the braking force, the car undergoes a uniformly accelerated motion with an initial speed of \(11.11 \mathrm{~m/s}\) and final speed 0 (as the car stops). Here, we use the equation of motion \( v^2 = u^2 + 2as \), where \(v\) is the final speed, \(u\) is the initial speed, \(a\) is acceleration and \(s\) is the distance covered. Re-arranging for \(s\), we get \(s_2 = \frac{v^2 - u^2}{2a} \). Thus, the distance covered during braking is \(s_2 = \frac{0^2 - (11.11 \mathrm{~m/s})^2}{2 \times -5.5 \mathrm{~m/s^2}} = 11.27 \mathrm{~m} \).
04

Determine Total Distance Travelled by Car

The total distance \(x_{\min}\) the car travels after the light changes before coming to a stop is the sum of the distance travelled during driver鈥檚 reaction time and the distance travelled during braking. So, \(x_{\min} = s_1 + s_2 = 8.33 \mathrm{~m} + 11.27 \mathrm{~m} = 19.6 \mathrm{~m}\).

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Key Concepts

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

Reaction Time
When driving, reaction time is crucial. It's the time it takes for a driver to perceive a signal and respond by starting to brake. In our scenario, the driver has a reaction time of 0.75 seconds. This period plays a significant role in determining the stopping distance of a vehicle. During reaction time, a vehicle travels at its original speed since the driver has not yet started to brake. For the car going at 11.11 m/s, the distance covered during this 0.75-second reaction phase is calculated easily. Simply multiply the speed by the reaction time, giving us a distance of approximately 8.33 meters. Understanding and accounting for reaction time helps in evaluating safe stopping distances and ensuring smooth transitions during driving and braking events.
Braking Distance
Once the driver starts braking, the car undergoes a different phase called braking distance. This is where the car transitions from its initial speed to a complete stop due to the deceleration brought about by braking forces.In this exercise, the car decelerates at a rate of -5.5 m/s虏. Using the equation of motion, we can calculate how much distance the car covers until it stops. The formula for this calculation is \[ s = \frac{v^2 - u^2}{2a} \]where \(s\) is the distance, \(v\) is the final velocity (0 m/s, since the car stops), \(u\) is the initial velocity (11.11 m/s), and \(a\) is the acceleration (-5.5 m/s虏). Plugging in these numbers, the braking distance works out to 11.27 meters. This information is vital for planning safe stopping spaces and understanding vehicle dynamics.
Uniformly Accelerated Motion
Uniformly accelerated motion is a fundamental concept in kinematics, dealing with objects moving with constant acceleration. This applies well in braking scenarios, where a car experiences steady deceleration until it stops. During braking, the car's speed decreases from an initial 11.11 m/s to 0 m/s at a uniform rate of -5.5 m/s虏. We utilize kinematic equations to determine the braking distance, ensuring that each component accurately contributes to calculating stopping distances. In uniformly accelerated motion, using equations of motion allows us to predict future positions based on known accelerations and time. It's helpful not just in studying vehicle dynamics, but also in understanding various real-world applications, such as sports and other physics-related phenomena. Applying these concepts helps in designing vehicles with effective braking systems and improving traffic safety overall.

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Most popular questions from this chapter

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