Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 81

A car moving at \(60.0 \mathrm{~km} / \mathrm{h}\) comes to a stop in \(t=4.00 \mathrm{~s}\) Assume uniform deceleration. a) How far does the car travel while stopping? b) What is its deceleration?

Short Answer

Expert verified
Question: A car is traveling at 60 km/h when the driver applies the brakes to come to a stop in 4 seconds. Calculate a) the distance the car travels while stopping, and b) the deceleration of the car. Answer: a) The car travels 33.4 meters while stopping. b) The deceleration of the car is -4.18 m/s².

Step by step solution

01

Convert the initial velocity to meters per second

We are given the initial velocity in km/h, but we'll need it in m/s to work with the equations of motion. To convert from km/h to m/s, we'll multiply by (1000 m/km) and divide by (3600 s/h): \(v_0 = 60.0 \frac{km}{h} \cdot \frac{1000 \thinspace m}{1 \thinspace km} \cdot \frac{1 \thinspace h}{3600 \thinspace s} = 16.7 \frac{m}{s}\)
02

Use the equations of motion to find the distance (a)

We'll use the equation \(v^2 = v_0^2 + 2ax\) to solve for the distance (x) the car travels while stopping. The final velocity (v) is 0 because the car comes to a stop: \(0 = (16.7 \thinspace m/s)^2 + 2a \cdot x\) Now, we'll solve for x: \(x = \frac{-(16.7 \thinspace m/s)^2}{2a}\) We still need to find the value of the deceleration (a) to calculate the distance, so let's proceed to the next step.
03

Use the equations of motion to find the deceleration (a)

We can use the equation \(v = v_0 + at\) to solve for the deceleration (a). We have the final velocity (v=0), the initial velocity (v₀=16.7 m/s), and the time (t=4.00 s): \(0 = 16.7 \frac{m}{s} + a(4.00 \thinspace s)\) Now we'll solve for a: \(a = -\frac{16.7 \frac{m}{s}}{4.00 \thinspace s} = -4.18 \frac{m}{s^2}\) Since the car is decelerating, the value of a is negative as expected.
04

Calculate the distance (a) using the found deceleration

Now that we have the deceleration (a), we can plug it back into the equation for the distance (x) we found in Step 2: \(x = \frac{-(16.7 \thinspace m/s)^2}{2(-4.18 \thinspace m/s^2)} = 33.4 \thinspace m\) So, the car travels 33.4 meters while stopping. In summary: a) The car travels 33.4 meters while stopping. b) The deceleration of the car is -4.18 m/s².

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The Bellagio Hotel in Las Vegas, Nevada, is well known for its Musical Fountains, which use 192 HyperShooters to fire water hundreds of feet into the air to the rhythm of music. One of the HyperShooters fires water straight upward to a height of \(240 \mathrm{ft}\). a) What is the initial speed of the water? b) What is the speed of the water when it is at half this height on its way down? c) How long will it take for the water to fall back to its original height from half its maximum height?

The position of a particle as a function of time is given as \(x(t)=\frac{1}{4} x_{0} e^{3 \alpha t}\), where \(\alpha\) is a positive constant. a) At what time is the particle at \(2 x_{0}\) ? b) What is the speed of the particle as a function of time? c) What is the acceleration of the particle as a function of time? d) What are the SI units for \(\alpha\) ?

An electron, starting from rest and moving with a constant acceleration, travels \(1.0 \mathrm{~cm}\) in \(2.0 \mathrm{~ms}\). What is the magnitude of this acceleration? a) \(25 \mathrm{~km} / \mathrm{s}^{2}\) b) \(20 \mathrm{~km} / \mathrm{s}^{2}\) c) \(15 \mathrm{~km} / \mathrm{s}^{2}\) d) \(10 \mathrm{~km} / \mathrm{s}^{2}\) e) \(5.0 \mathrm{~km} / \mathrm{s}^{2}\)

You are flying on a commercial airline on your way from Houston, Texas, to Oklahoma City, Oklahoma. Your pilot announces that the plane is directly over Austin, Texas, traveling at a constant speed of \(245 \mathrm{mph}\), and will be flying directly over Dallas, Texas, \(362 \mathrm{~km}\) away. How long will it be before you are directly over Dallas, Texas?

Two athletes jump straight up. Upon leaving the ground, Adam has half the initial speed of Bob. Compared to Adam, Bob jumps a) 0.50 times as high. b) 1.41 times as high. c) twice as high. d) three times as high. e) four times as high.
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Translational Dynamics

Read Explanation

Measurements

Read Explanation

Classical Mechanics

Read Explanation

Wave Optics

Read Explanation

Conservation of Energy and Momentum

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free