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Chapter 2: Problem 73

A person driving her car at 35 km/h approaches an intersection just as the traffic light turns yellow. She knows that the yellow light lasts only 2.0 s before turning to red, and she is 28 m away from the near side of the intersection (Fig. 2-49). Should she try to stop, or should she speed up to cross the intersection before the light turns red? The intersection is 15 m wide. Her car's maximum deceleration is \(-\)5.8 m/s\(^2\), whereas it can accelerate from 45 km/h to 65 km/h in 6.0 s. Ignore the length of her car and her reaction time.

Short Answer

Expert verified
She should stop before the intersection.

Step by step solution

01

Convert Speed from km/h to m/s

First, we need to convert the initial speed of the car from km/h to m/s. The car is moving at 35 km/h. To convert this to m/s, we use the conversion factor: \[1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}\]Thus, \[35 \text{ km/h} = \frac{35}{3.6} \approx 9.72 \text{ m/s}\]This is the car's initial speed in meters per second.
02

Calculate Stopping Distance

Next, we calculate whether the car can stop in time before the intersection. The formula for stopping distance when the maximum deceleration is given is: \[v^2 = u^2 + 2as\]where:- \(v = 0\) m/s (final velocity, as the car stops)- \(u = 9.72\) m/s (initial speed)- \(a = -5.8\) m/s ext{虏} (deceleration)- \(s\) is the stopping distanceRearrange to solve for \(s\):\[0 = (9.72)^2 + 2(-5.8)s \2(5.8)s = (9.72)^2 \s = \frac{(9.72)^2}{2 \times 5.8} \approx 8.15 \text{ m}\]The stopping distance is approximately 8.15 m, which is less than 28 m to the intersection. Therefore, she can stop in time.
03

Determine If Car Can Accelerate Through Intersection

We now check if she can speed through the intersection. First, calculate the time it takes to clear the intersection with the maximum acceleration.Given the car's acceleration capability from 45 km/h to 65 km/h in 6 seconds, first convert those speeds to m/s:\[45 \text{ km/h} = \frac{45}{3.6} \approx 12.5 \text{ m/s}\]\[65 \text{ km/h} = \frac{65}{3.6} \approx 18.06 \text{ m/s}\]Acceleration \(a\) can be calculated as:\[a = \frac{18.06 \text{ m/s} - 12.5 \text{ m/s}}{6 \text{ s}} \approx 0.926 \text{ m/s}^2\]Using the initial speed of 9.72 m/s, find the distance covered in 2 seconds:\[ s = ut + \frac{1}{2}at^2 \s = 9.72 \times 2 + \frac{1}{2} \times 0.926 \times (2)^2 \s = 19.44 + 1.852 \approx 21.29 \text{ m} \]Since 21.29 m is less than the 28 m to the intersection and barely covers it (incorrect because 15 m requires more coverage), she can't clear the intersection in time.
04

Conclusion: Choose the Best Option

Since she can stop before reaching the intersection (only requiring 8.15 m but having 28 m to stop), and she cannot pass through before it turns red (only travels approximately 21.29 m in 2 s with acceleration), the better option is to stop.

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Key Concepts

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

Deceleration
Deceleration is the process of slowing down and reducing velocity over time. In essence, it's negative acceleration. This happens when a force acts opposite to the direction of motion. For example, when a car comes to a stop, it experiences deceleration because the brakes exert force opposite its movement.

In kinematics, deceleration is calculated using the formula:
  • Final velocity ( \( v \) ) becomes zero as the car stops
  • Initial velocity ( \( u \) ) is the speed before deceleration starts
  • Acceleration ( \( a \) ) is a negative value because the velocity is decreasing
The stopping distance can be determined using the equation:\[v^2 = u^2 + 2as\]where \( s \) is the stopping distance. In scenarios like approaching an intersection, knowing the deceleration helps determine if stopping before a crosswalk is feasible.

Deceleration is crucial for ensuring safety on the road. It is necessary to evaluate whether a car can halt promptly and safely within the available distance.
Acceleration
Acceleration refers to the rate of change of velocity over time. It's what allows a vehicle to increase speed. When the driver decides to speed up to clear an intersection, this concept becomes critical.

To calculate acceleration, use the formula:\[a = \frac{\Delta v}{t}\]Where:
  • \( \Delta v \) is the change in velocity
  • \( t \) is the time over which acceleration occurs
A positive acceleration value means the car is speeding up. For instance, in the context of the original problem, this means transitioning from an initial speed to a higher speed within a specific duration.

Understanding how quickly a vehicle can accelerate is crucial when making decisions, such as crossing an intersection. If a vehicle can't accelerate enough in a given time, it might not safely pass through before a light changes.
Stopping Distance
Stopping distance is the total length a vehicle travels from the time a driver reacts to something up ahead until the vehicle comes to a complete stop. It's important to ensure a vehicle can stop within a certain distance to avoid accidents, especially at intersections.

Stopping distance comprises two parts:
  • Reaction distance: The distance covered by a car during the driver's reaction time.
  • Braking distance: The distance the car travels while slowing down to a stop.
In the given problem, the initial braking distance is calculated by using the formula:\[s = \frac{(\text{initial speed})^2}{2 \times \text{deceleration magnitude}}\]The calculated stopping distance allows the driver to assess whether they can stop safely before an intersection. Given the parameters, it's clear that the car can halt within 8.15 m, well before reaching the intersection.

Understanding stopping distance is an essential aspect of driving and road safety. Knowing how quickly and effectively a vehicle can be brought to a halt ensures informed decisions on the road, preventing potential collisions.

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Most popular questions from this chapter

Agent Bond is standing on a bridge, 15 m above the road below, and his pursuers are getting too close for comfort. He spots a flatbed truck approaching at 25 m/s, which he measures by knowing that the telephone poles the truck is passing are 25 m apart in this region. The roof of the truck is 3.5 m above the road, and Bond quickly calculates how many poles away the truck should be when he drops down from the bridge onto the truck, making his getaway. How many poles is it?

A conveyor belt is used to send burgers through a grilling machine. If the grilling machine is 1.2 m long and the burgers require 2.8 min to cook, how fast must the conveyor belt travel? If the burgers are spaced 25 cm apart, what is the rate of burger production (in burgers/min)?

(II) Digital bits on a 12.0-cm diameter audio CD are encoded along an outward spiraling path that starts at radius \(R_1 =\) 2.5 cm and finishes at radius \(R_2 =\) 5.8 cm. The distance between the centers of neighboring spiralwindings is 1.6 \(\mu\)m (= 1.6 \(\times\) 10\(^{-6}\) m). (\(a\)) Determine the total length of the spiraling path. [\(Hint\): Imagine "unwinding" the spiral into a straight path of width 1.6\(\mu\)m and note that the original spiral and the straight path both occupy the same area.] (\(b\)) To read information, a CD player adjusts the rotation of the CD so that the player's readout laser moves along the spiral path at a constant speed of about 1.2 m/s. Estimate the maximum playing time of such a CD.

Pelicans tuck their wings and free-fall straight down when diving for fish. Suppose a pelican starts its dive from a height of 14.0 m and cannot change its path once committed. If it takes a fish 0.20 s to perform evasive action, at what minimum height must it spot the pelican to escape? Assume the fish is at the surface of the water.

(III) A falling stone takes 0.31 s to travel past a window 2.2 m tall (Fig. 2-41). From what height above the top of the window did the stone fall?
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