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Chapter 2: Problem 39

A car traveling at \(25 \mathrm{mi} / \mathrm{h}\) is to stop on a \(35-\mathrm{m}\) -long shoulder of the road. (a) What is the required magnitude of the minimum acceleration? (b) How much time will elapse during this minimum deceleration until the car stops?

Short Answer

Expert verified
(a) The required deceleration is \( -1.784 \text{ m/s}^2 \). (b) The car stops in 6.27 seconds.

Step by step solution

01

Identify Known Information

We have the initial velocity \( v_i = 25 \text{ mi/h} \). Since most scientific calculations are conducted in metric units, we'll convert this first to \( \text{m/s} \). Given that \( 1 \text{ mi} = 1609.34 \text{ m} \) and \( 1 \text{ h} = 3600 \text{ s} \), we use:\[ v_i = 25 \frac{\text{mi}}{\text{h}} \times \frac{1609.34 \text{ m}}{1 \text{ mi}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 11.176 \text{ m/s} \]The stopping distance \( s = 35 \text{ m} \), and the final velocity \( v_f = 0 \text{ m/s} \) as the car comes to a stop.
02

Use Uniform Acceleration Equation

Identify the right formula to find the acceleration. We use the equation for motion with uniform acceleration:\[ v_f^2 = v_i^2 + 2a \cdot s \]Rearranging to solve for acceleration \( a \):\[ a = \frac{v_f^2 - v_i^2}{2s} \]
03

Substitute Known Values

Substitute the known values into the equation to find \( a \):\[ a = \frac{0 - (11.176)^2}{2 \times 35} \]Calculate:\[ a = \frac{-124.891}{70} = -1.784 \text{ m/s}^2 \]
04

Interpret Result for Part (a)

The acceleration needs to be \( a = -1.784 \text{ m/s}^2 \). The negative sign indicates deceleration, meaning slowing down.
05

Use Deceleration to Find Time

To find the time \( t \) taken to stop, use the formula:\[ v_f = v_i + a \cdot t \]Solve for \( t \):\[ 0 = 11.176 + (-1.784) \cdot t \]\[ t = \frac{-11.176}{-1.784} \]
06

Calculate Time

Calculate \( t \):\[ t = \frac{11.176}{1.784} = 6.27 \text{ seconds} \]
07

Interpret Result for Part (b)

The car takes approximately 6.27 seconds to come to a stop.

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Key Concepts

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

Uniform Acceleration
Uniform acceleration occurs when an object changes its velocity at a constant rate. It's straightforward because all changes in speed happen evenly over time. This means that the increase or decrease in velocity is always the same each second.

For example, if a car accelerates uniformly, it might increase its speed by 2 m/s every second. This results in a predictable change in motion. It鈥檚 like walking up a staircase a step at a consistent pace.
  • Constant Rate: The rate of change in velocity is the same in every equal time interval.
  • Predictable Motion: By knowing the acceleration, you can predict where the object will be after a certain time.
  • Mathematical Representation: This can be expressed with straightforward equations which make calculations simpler.
In scenarios such as stopping a vehicle, uniform acceleration (or more often deceleration) helps calculate necessary distances and times to reach desired speeds.
Deceleration
Deceleration is simply negative acceleration. It means slowing down. When a vehicle decelerates, its speed decreases over time due to a negative acceleration value. In our example, the car鈥檚 deceleration is shown by the negative sign in the calculated acceleration. This shows the car is reducing its speed to come to a stop.

Let's break down deceleration more:
  • Negative Acceleration: Represented by a negative number in calculations, showing a reduction in speed.
  • Stopping Distance: Knowing deceleration helps us find out how much space is needed to stop safely.
  • Calculation Formula: In uniform deceleration, we use similar formulas as acceleration, but outcomes indicate a decrease rather than increase in speed.
Therefore, understanding deceleration is crucial for safety and efficiency in situations where stopping is necessary.
Kinematic Equations
Kinematic equations are vital tools in physics for describing motion. They help us calculate things like velocity, acceleration, time, and displacement without needing to know everything about the forces involved.

The two primary equations showcased in our exercise include:
  • For acceleration: \( v_f^2 = v_i^2 + 2a \cdot s \), where we solve for acceleration \(a\) using given velocities and displacement.
  • For time: \( v_f = v_i + a \cdot t \), used to determine the time taken for the car to come to rest.
These equations bring clarity and simplicity:
  • General Application: They apply to many different situations involving linear motion.
  • Substitution: Known variables are plugged into the formulas to solve for unknowns.
  • Predictive Power: They allow us to predict future positions or speeds given current conditions.
Understanding and using kinematic equations enables students to handle complex motion problems with greater ease.

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Most popular questions from this chapter

A hockey puck sliding along the ice to the left hits the boards head-on with a speed of \(35 \mathrm{~m} / \mathrm{s}\). As it reverses direction, it is in contact with the boards for \(0.095 \mathrm{~s}\), before rebounding at a slower speed of \(11 \mathrm{~m} / \mathrm{s}\). Determine the average acceleration the puck experienced while hitting the boards. Typical car accelerations are \(5.0 \mathrm{~m} / \mathrm{s}^{2} .\) Comment on the size of your answer, and why it is so different from this value, especially when the puck speeds are similar to car speeds.

Fig. \(2.25,\) a student at a window on the second floor of a dorm sees his math professor walking on the sidewalk beside the building. He drops a water balloon from \(18.0 \mathrm{~m}\) above the ground when the professor is \(1.00 \mathrm{~m}\) from the point directly beneath the window. If the professor is \(1.70 \mathrm{~m}\) tall and walks at a rate of \(0.450 \mathrm{~m} / \mathrm{s},\) does the balloon hit her? If not, how close does it come?

An automobile traveling at \(15.0 \mathrm{~km} / \mathrm{h}\) along a straight, level road accelerates to \(65.0 \mathrm{~km} / \mathrm{h}\) in \(6.00 \mathrm{~s}\). What is the magnitude of the auto's average acceleration?

On a water slide ride, you start from rest at the top of a 45.0 -m-long incline (filled with running water) and accelerate down at \(4.00 \mathrm{~m} / \mathrm{s}^{2}\). You then enter a pool of water and skid along the surface for \(20.0 \mathrm{~m}\) before stopping. (a) What is your speed at the bottom of the incline? (b) What is the deceleration caused by the water in the pool? (c) What was the total time for you to stop? (d) How fast were you moving after skidding the first \(10.0 \mathrm{~m}\) on the water surface?

A pollution-sampling rocket is launched straight upward with rockets providing a constant acceleration of \(12.0 \mathrm{~m} / \mathrm{s}^{2}\) for the first \(1000 \mathrm{~m}\) of flight. At that point the rocket motors cut off and the rocket itself is in free fall. Ignore air resistance. (a) What is the rocket's speed when the engines cut off? (b) What is the maximum altitude reached by this rocket? (c) What is the time it takes to get to its maximum altitude?
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