/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 48 You are driving to the grocery s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Q. 48 (page 58)

You are driving to the grocery store at 20 m/s. You are 110 m

from an intersection when the traffic light turns red. Assume

that your reaction time is 0.50 s and that your car brakes with

constant acceleration. What magnitude braking acceleration will bring you to a stop exactly at the intersection?

Short Answer

Expert verified

The magnitude of breaking acceleration to bring the car at a rest is2m/s2.

Step by step solution

01

Given information

Initial velocity u=20m/s, Reaction time t=0.5s, Distance of intersection,d=110m.

02

Calculation

The distance travelled after pressing the brake is, s=uts=20脳0.5ms=10m

So the distance traversed before pressing the brake is, D=110-10m=100m.

03

Find the magnitude of acceleration

We can find the acceleration by using the formula,

v2=u2+2Da, where v鈫�finalvelocity,a鈫�acceleration.

The final velocity of the car should be, v=0at the time of stopping.

Therefore, the acceleration to bring the car at a stop can be determined as,

0=u2+2a脳100a=-u2200a=-202200a=-2m/s2

Therefore the magnitude of acceleration to bring the car at a rest is 2m/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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

FIGURE EX2.31 shows the acceleration-versus-time graph of a particle moving along the x-axis. Its initial velocity is v0x = 8.0 m/s at t0 = 0 s. What is the particle鈥檚 velocity at t = 4.0 s?

You are 9.0 m from the door of your bus, behind the bus, when it pulls away with an acceleration of 1.0 m/s2. You instantly start running toward the still-open door at 4.5 m/s.

a. How long does it take for you to reach the open door and

jump in?

b. What is the maximum time you can wait before starting to run and still catch the bus?

A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at a height h directly above the first ball is dropped from rest.
a. At what height above the ground do the balls collide? Your answer will be an algebraic expression in terms of h, v0, and g.
b. What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?
c. For what value of h does the collision occur at the instant when the first ball is at its highest point?

A 1000 kg weather rocket is launched straight up. The rocket

motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance. What was the rocket鈥檚 acceleration during the first 16 s?

Larry leaves home at 9:05 and runs at a constant speed to the lamp post seen in FIGURE EX2.3. He reaches the lamppost at 9:07, immediately turns, and runs to the tree. Larry arrives at the tree at 9:10. a. What is Larry鈥檚 average velocity, in m/min, during each of these two intervals? b. What is Larry鈥檚 average velocity for the entire run?
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Fields in Physics

Read Explanation

Waves Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Space Physics

Read Explanation

Electricity and Magnetism

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.