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You are driving to the grocery store at \(20 \mathrm{m} / \mathrm{s}\). You are \(110 \mathrm{m}\) from an intersection when the traffic light turns red. Assume that your reaction time is \(0.50 \mathrm{s}\) and that your car brakes with constant acceleration. a. How far are you from the intersection when you begin to apply the brakes? b. What acceleration will bring you to rest right at the intersection? c. How long does it take you to stop after the light turns red?

Short Answer

Expert verified
a. You are \(10 \mathrm{m}\) from the intersection when beginning to apply brakes. b. The required acceleration (or deceleration) is \(-2 \mathrm{m} / \mathrm{s}^{2}\). c. It takes \(10.5 \mathrm{s}\) to stop after the light turns red.

Step by step solution

01

Calculate the distance travelled during the reaction time

Firstly, the distance travelled during the reaction time before applying the brakes is calculated by multiplying the speed (\(20 \mathrm{m} / \mathrm{s}\)) with the reaction time (\(0.50 \mathrm{s}\)). So, distance = speed * time = \(20 \mathrm{m} / \mathrm{s}\) * \(0.50 \mathrm{s}\) = \(10 \mathrm{m}\)
02

Calculate the required acceleration

After determining the distance in step 1, subtract it from the total distance to the intersection to find the distance over which the car must decelerate. This is \(110 \mathrm{m}\) - \(10 \mathrm{m}\) = \(100 \mathrm{m}\). Now using the second equation of motion, which is \(v^{2} = u^{2} + 2as\), where final velocity v = 0 (as car comes to rest), initial velocity u = \(20 \mathrm{m} / \mathrm{s}\), distance s = \(100 \mathrm{m}\), we solve for a (acceleration). Rearranging the equation gives a = \(-(u^{2}) / (2s)\) = \(-((20 \mathrm{m} / \mathrm{s})^{2}) / (2 * 100 \mathrm{m})\) = \(-2 \mathrm{m} / \mathrm{s}^{2}\). The negative sign indicates deceleration.
03

Calculate the total stopping time

The total time to stop the car is the sum of the reaction time and the time needed to stop the car while braking. The time during braking is calculated using the first equation of motion, which is \(v = u + at\), where final velocity v = 0, initial velocity u = \(20 \mathrm{m} / \mathrm{s}\), acceleration a = \(-2 \mathrm{m} / \mathrm{s}^{2}\). Solving for t (time), t = \((v - u) / a\) = \((0 - 20 \mathrm{m} / \mathrm{s}) / \(-2 \mathrm{m} / \mathrm{s}^{2}\)\) = \(10 \mathrm{s}\). Adding reaction time, total time = \(0.5 \mathrm{s}\) + \(10 \mathrm{s}\) = \(10.5 \mathrm{s}\).

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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 occurs when an object's velocity changes at a steady rate over time. For example, when you apply the brakes on a car, there is a uniform increase (or in the case of slowing down, decrease) in the rate of speed, which is acceleration.

This concept is central in calculating various aspects of motion, such as the distance covered or the velocity reached over a certain period. Constant acceleration can be either positive, leading to an increase in speed, or negative, which is commonly referred to as deceleration. Negative acceleration is crucial when discussing braking distances and traffic safety.

The ability to calculate the effects of constant acceleration allows drivers and automated systems to understand how quickly a vehicle can stop or reach a certain speed, making it a fundamental concept for ensuring safe driving practices.
Equations of Motion
The equations of motion are mathematical formulas that describe the relationships between velocity, acceleration, distance, and time when acceleration is constant. These equations enable us to predict an object's future position and velocity or to back-calculate conditions from known outcomes, such as in the case of a car approaching a red light.

They encompass three primary formulas:
  • \(v = u + at\) where 'v' is the final velocity, 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time elapsed.
  • \(s = ut + \frac{1}{2}at^2\) where 's' is the distance traveled.
  • \(v^2 = u^2 + 2as\) which is particularly useful for problems involving starting and stopping, as it links velocity and distance without the direct need for time.

These equations are the stepping stones for solving kinematics problems in physics, including those involving vehicles in traffic.
Deceleration
Deceleration is a specific type of acceleration that occurs when an object slows down, resulting in a negative acceleration value. In our driving scenario, deceleration is paramount in bringing the car to a safe and complete stop at the red traffic light.

It is calculated using the equations of motion and the known values of initial speed, final speed (often zero in stopping scenarios), and the distance over which the car must stop. Understanding and correctly calculating deceleration enables drivers to respond appropriately to traffic signals and helps in designing failsafe measures for automated vehicles, ensuring safer driving conditions for everyone on the road.
Reaction Time
Reaction time plays an integral role in driving and in solving kinematics problems. It is the period between the perception of a signal (such as a red light) and the physical response (applying the brakes).

In our exercise, the driver's reaction time of 0.50 seconds contributes to the total distance needed to stop the car. During this half a second, the car continues to travel at its initial speed before the brakes are even applied, thus extending the stopping distance.

Understanding the impact of reaction time is key to both driving safety and physics calculations. It highlights the importance of staying alert and prepared to respond because sometimes a fraction of a second can make a significant difference in avoiding collisions.

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Most popular questions from this chapter

David is driving a steady \(30 \mathrm{m} / \mathrm{s}\) when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady \(2.0 \mathrm{m} / \mathrm{s}^{2}\) at the instant when David passes. a. How far does Tina drive before passing David? b. What is her speed as she passes him?

One game at the amusement park has you push a puck up a long, frictionless ramp. You win a stuffed animal if the puck, at its highest point, comes to within \(10 \mathrm{cm}\) of the end of the ramp without going off. You give the puck a push, releasing it with a speed of \(5.0 \mathrm{m} / \mathrm{s}\) when it is \(8.5 \mathrm{m}\) from the end of the ramp. The puck's speed after traveling \(3.0 \mathrm{m}\) is \(4.0 \mathrm{m} / \mathrm{s}\). Are you a winner?

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.0 s later. What was the rocket's acceleration?

A driver has a reaction time of \(0.50 \mathrm{s}\), and the maximum deceleration of her car is \(6.0 \mathrm{m} / \mathrm{s}^{2} .\) She is driving at \(20 \mathrm{m} / \mathrm{s}\) when suddenly she sees an obstacle in the road \(50 \mathrm{m}\) in front of her. Can she stop the car in time to avoid a collision?

A car starts from rest at a stop sign. It accelerates at \(4.0 \mathrm{m} / \mathrm{s}^{2}\) for \(6.0 \mathrm{s},\) coasts for \(2.0 \mathrm{s},\) and then slows down at a rate of \(3.0 \mathrm{m} / \mathrm{s}^{2}\) for the next stop sign. How far apart are the stop signs?

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