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

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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

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College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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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

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

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