2)A 1300-kg sports car initially has a velocity of 35.2 m/s due south. The driver slams the brakes to stop over a 190 m distance. a) What is the magnitude of tihe car's acceleration, in m/s? ? 1300k aニー 36.13 m/ş? %3D 35.2 e/s b) What average net force magnitude was necessary to stop the car? 1300 (35.2m/3)². Fnet = ニ4,29N 2(190 m) c) Assuming the tires do not skid, what coefficient of static firiction between the tires and pavement is meeded? = xN =36.93m/2)(4.24 N) -15C. 5N %3D
2)A 1300-kg sports car initially has a velocity of 35.2 m/s due south. The driver slams the brakes to stop over a 190 m distance. a) What is the magnitude of tihe car's acceleration, in m/s? ? 1300k aニー 36.13 m/ş? %3D 35.2 e/s b) What average net force magnitude was necessary to stop the car? 1300 (35.2m/3)². Fnet = ニ4,29N 2(190 m) c) Assuming the tires do not skid, what coefficient of static firiction between the tires and pavement is meeded? = xN =36.93m/2)(4.24 N) -15C. 5N %3D
College Physics
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Chapter1: Units, Trigonometry. And Vectors
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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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Solve #2, includes: a, b, and c.
![### Physics Problem: Car Braking
A 1300-kg sports car initially has a velocity of 35.2 m/s due south. The driver slams the brakes to stop over a 190 m distance.
**a) What is the magnitude of the car’s acceleration, in m/s²?**
\[ a = \frac{1300 \times 35.2^2}{2 \times 190} = 36.73 \, \text{m/s}^2 \]
**b) What average net force magnitude was necessary to stop the car?**
\[ F_{\text{net}} = \frac{1300 \times (35.2)^2}{2(190)} = 4.24 \, \text{N} \]
**c) Assuming the tires do not skid, what coefficient of static friction between the tires and pavement is needed?**
\[ f_s = \alpha \times N \]
\[ = \frac{(36.93 \, \text{m/s}^2)(4.24 \, \text{N})}{156.58 \, \text{N}} \]
The equations show the steps taken to solve for acceleration, force, and friction, with values substituted for the given problem scenario.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd9ac933e-a222-4210-8a67-76a29c4a48c0%2F2f19a7fb-f358-4936-bd3b-58219c599a47%2Fb0ibwl_processed.png&w=3840&q=75)
Transcribed Image Text:### Physics Problem: Car Braking
A 1300-kg sports car initially has a velocity of 35.2 m/s due south. The driver slams the brakes to stop over a 190 m distance.
**a) What is the magnitude of the car’s acceleration, in m/s²?**
\[ a = \frac{1300 \times 35.2^2}{2 \times 190} = 36.73 \, \text{m/s}^2 \]
**b) What average net force magnitude was necessary to stop the car?**
\[ F_{\text{net}} = \frac{1300 \times (35.2)^2}{2(190)} = 4.24 \, \text{N} \]
**c) Assuming the tires do not skid, what coefficient of static friction between the tires and pavement is needed?**
\[ f_s = \alpha \times N \]
\[ = \frac{(36.93 \, \text{m/s}^2)(4.24 \, \text{N})}{156.58 \, \text{N}} \]
The equations show the steps taken to solve for acceleration, force, and friction, with values substituted for the given problem scenario.
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