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Chapter 2: Problem 40

You are driving toward a traffic signal when it turns yellow. Your speed is the legal speed limit of \(v_{0}=55 \mathrm{~km} / \mathrm{h}\); your best deceleration rate has the magnitude \(a=5.18 \mathrm{~m} / \mathrm{s}^{2}\). Your best reaction time to begin braking is \(T=0.75 \mathrm{~s}\). To avoid having the front of your car enter the intersection after the light turns red, should you brake to a stop or continue to move at \(55 \mathrm{~km} / \mathrm{h}\) if the distance to the intersection and the duration of the yellow light are (a) \(40 \mathrm{~m}\) and \(2.8 \mathrm{~s}\), and (b) \(32 \mathrm{~m}\) and \(1.8 \mathrm{~s}\) ? Give an answer of brake, continue, either (if either strategy works), or neither (if neither strategy works and the yellow duration is inappropriate).

Short Answer

Expert verified
(a) Either, (b) Neither.

Step by step solution

01

Convert Speed to Meters per Second

First, convert the initial speed from kilometers per hour to meters per second. Use the conversion factor: \[ v_0 = 55 \text{ km/h} \times \frac{1000 \text{ m/km}}{3600 \text{ s/h}} = 15.28 \text{ m/s} \]
02

Calculate Braking Distance

Next, calculate the distance required to stop the car after the reaction time. Using the formula: \[ d = v_0 \cdot T + \frac{v_0^2}{2a} \]- Reaction distance: \( d_{\text{reaction}} = 15.28 \times 0.75 = 11.46 \text{ m} \)- Braking distance: \( d_{\text{braking}} = \frac{15.28^2}{2 \times 5.18} = 22.54 \text{ m} \)Total stopping distance: \( d_{\text{stop}} = 11.46 + 22.54 = 34 \text{ m} \)
03

Evaluate Situation (a)

With a distance to the intersection of 40 m and a yellow light duration of 2.8 s:- Stopping distance (34 m) is less than the distance to the intersection (40 m), so stopping before the intersection is possible.- Check if continuing is feasible by calculating if the car can cross 40 m in 2.8 s:\[ \text{Time to cross} = \frac{40}{15.28} = 2.62 \text{ s} \]2.62 s < 2.8 s, so either action is possible. Answer: either.
04

Evaluate Situation (b)

With a distance to the intersection of 32 m and yellow light duration of 1.8 s:- Stopping distance (34 m) is greater than the distance to the intersection (32 m), so stopping is not possible.- Check if continuing with a speed of 15.28 m/s will pass the intersection in 1.8 s:\[ \text{Time to cross} = \frac{32}{15.28} = 2.09 \text{ s} \]2.09 s > 1.8 s, so continuing will not avoid the red light. Answer: neither.

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Key Concepts

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

Deceleration
Deceleration refers to the negative acceleration of an object, which means that the object is slowing down. In the context of driving, deceleration is crucial because it determines how quickly a car can come to a stop. It is often measured in meters per second squared (m/s虏).
Understanding deceleration is key when calculating the stopping distance of a vehicle. The stopping distance is influenced by both the initial speed of the vehicle and its deceleration rate. A higher deceleration rate means the car will stop in a shorter distance.
In our exercise, the car's deceleration rate is given as \( a = 5.18 \, \text{m/s}^2 \). This value is used to compute the braking distance after the driver begins to decelerate.
  • Deceleration indicates how rapidly a vehicle slows down.
  • It's an essential factor in safe stopping, especially at traffic signals.
  • It is used alongside initial speed and reaction time to determine total stopping distance.
To sum up, knowing your car's deceleration ability can help you predict and prepare your stops effectively, especially when approaching intersections.
Reaction Time
Reaction time is the period it takes for a driver to respond to a stimulus, such as a traffic light changing color. During reaction time, a vehicle continues to travel at its initial speed since the driver hasn't yet applied the brakes.
In this exercise, the driver's best reaction time is \( T = 0.75 \, \text{s} \). This must be factored in when estimating the total stopping distance since it directly affects how far the vehicle travels before the brakes engage.
During these \(0.75\) seconds, while the driver is reacting:
  • The vehicle covers a distance known as the reaction distance.
  • Until braking starts, this distance is determined by the speed of the vehicle.
This concept is vital in safe driving practices, emphasizing the importance of remaining alert and reducing distractions while driving to minimize reaction time.
Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces causing the motion. It primarily involves the study of velocity, acceleration, and displacement.
In traffic situations, kinematics helps predict whether a vehicle can stop in time or will pass an intersection before the light changes. This involves calculating motions based on initial speed, acceleration, and time.
  • Kinematics equations help determine distance and time relationships for moving vehicles.
  • They provide the mathematical framework to compute stopping distances and times to cross an intersection.
In our exercise, the kinematics principles applied include:
1. Converting speed to meters per second for consistent units.
2. Calculating both reaction and braking distances to predict whether stopping in time is possible based on the initial conditions. By understanding kinematics, drivers can anticipate their vehicle's movement in response to signals, ensuring safe driving decisions during varying traffic conditions.

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Most popular questions from this chapter

The position of a particle moving along the \(x\) axis is given in centimeters by \(x=9.75+1.50 t^{3}\), where \(t\) is in seconds. Calculate (a) the average velocity during the time interval \(t=2.00 \mathrm{~s}\) to \(t=3.00 \mathrm{~s}\); (b) the instantaneous velocity at \(t=2.00 \mathrm{~s} ;\) (c) the instantaneous velocity at \(t=3.00 \mathrm{~s} ;\) (d) the instantaneous velocity at \(t=2.50 \mathrm{~s} ;\) and (e) the instantaneous velocity when the particle is midway between its positions at \(t=2.00 \mathrm{~s}\) and \(t=3.00 \mathrm{~s}\). (f) Graph \(x\) versus \(t\) and indicate your answers graphically.

The brakes on your car can slow you at a rate of \(5.2 \mathrm{~m} / \mathrm{s}^{2}\). (a) If you are going \(146 \mathrm{~km} / \mathrm{h}\) and suddenly see a state trooper, what is the minimum time in which you can get your car under the \(90 \mathrm{~km} / \mathrm{h}\) speed limit? (The answer reveals the futility of braking to keep your high speed from being detected with a radar or laser gun.) (b) Graph \(x\) versus \(t\) and \(v\) versus \(t\) for such a slowing.

An electron, starting from rest and moving with a constant acceleration, travels \(2.00 \mathrm{~cm}\) in \(5.00 \mathrm{~ms}\). What is the magnitude of this acceleration?

A certain elevator cab has a total run of \(190 \mathrm{~m}\) and a maximum speed of \(305 \mathrm{~m} / \mathrm{min}\), and it accelerates from rest and then back to rest at \(1.22 \mathrm{~m} / \mathrm{s}^{2}\). (a) How far does the cab move while accelerating to full speed from rest? (b) How long does it take to make the nonstop \(190 \mathrm{~m}\) run, starting and ending at rest?

Catapulting mushrooms. Certain mushrooms launch their spores by a catapult mechanism. As water condenses from the air onto a spore that is attached to the mushroom, a drop grows on one side of the spore and a film grows on the other side. The spore is bent over by the drop's weight, but when the film reaches the drop, the drop's water suddenly spreads into the film and the spore springs upward so rapidly that it is slung off into the air. Typically, the spore reaches a speed of \(1.6 \mathrm{~m} / \mathrm{s}\) in a \(5.0 \mu \mathrm{m}\) launch; its speed is then reduced to zero in \(1.0 \mathrm{~mm}\) by the air. Using those data and assuming constant accelerations, find the acceleration in terms of \(g\) during (a) the launch and (b) the speed reduction.
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