3. A 2800-lb car enters an S-curve at A with a speed of 45 mi/hr with brakes applied to reduce the speedto 25 mi/hr at a uniform rate in a distance of 300 ft measured along the curve from A to B. The radiusof curvature at B is 600 feet. Calculate the total friction force exerted by the road on the tires at B.The road is in a horizontal plane.B300600

Name: 3. A 2800-lb car enters an S-curve at A with a speed of 45 mi/hr with
Uploaded: 2024-03-20T06:56:46.000Z
Channel: Bartleby
Description: 3. A 2800-lb car enters an S-curve at A with a speed of 45 mi/hr with brakes applied to reduce the speedto 25 mi/hr at a uniform rate in a distance of 300 ft measured along the curve from A to B. The radiusof curvature at B is 600 feet. Calculate th

Answered: 3. A 2800-lb car enters an S-curve at A…

Physics for Scientists and Engineers with Modern Physics

10th Edition

ISBN:9781337553292

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter6: Circular Motion And Other Applications Of Newton's Laws

Section: Chapter Questions

Problem 44AP: A model airplane of mass 0.750 kg flies with a speed of 35.0 m/s in a horizontal circle at the end...

See similar textbooks

Related questions

Concept explainers

Topic Video

Question

3. A 2800-lb car enters an S-curve at A with a speed of 45 mi/hr with brakes applied to reduce the speed
to 25 mi/hr at a uniform rate in a distance of 300 ft measured along the curve from A to B. The radius
of curvature at B is 600 feet. Calculate the total friction force exerted by the road on the tires at B.
The road is in a horizontal plane.
B
300
600

Expert Solution

Step 1

mass of the car m = 2800 lb = 1270.059 Kg

initial velocity of car u = 45 mi/hr = 20.1168 m/s

final velocity of car v = 25 mi/hr = 11.176 m/s

Distance s = 300 ft = 91.44 m

Radius of curvature= 600 ft $ρ = 182.88 m$

We know from the third equation of motion calculating acceleration from A to B in tangential direction

$v^{2} = u^{2} + 2 a_{t} s 11 . 176^{2} = 20 . 1168^{2} + 2 \times a_{t} \times 91.44 124.90 = 404.68 + 182.88 a_{t} a_{t} = \frac{124.90 - 404.68}{182.88} a_{t} = - 1.53 m / s^{2}$

Now we are calculating acceleration of car at normal

$a_{n} = \frac{v^{2}}{ρ} a_{n} = \frac{u^{2}}{ρ} a_{n} = \frac{20 . 1168^{2}}{182.88} a_{n} = 2.21 m / s^{2}$

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions