The second photo has the background information needed to solve the practice problem. **Practice Problem 4.4:**If passenger comfort requires that the acceleration should be no greater in magnitude than 0.10g, what distance is required to stop the car if its speed is initially 30 m/s (roughly 60 mi/h)? What force is required?**Answers:** 459 m, 1960 N. **Example 4.9: Quick Stop of a Heavy Car**In this example, we consider a problem with the weight of an object. In the x-direction, the force that makes the car stop is the force of kinetic friction. We use a modern luxury car weighing 19.6 x 10³N (about 4400 lb) traveling in the +x direction making a fast stop. The x-component of the net force acting on it is -1.50 x 10⁴N. What is its acceleration?**Solution****Set Up**Figure 4.19 shows our diagram for this problem. We use the weight as force and to find acceleration of the object. For acceleration, proceeding to use Newton's second law is necessary. For completeness, include the force of equal magnitude on the car in the y-direction.**Solve**To find acceleration, we use Newton’s second law. First, however, we need to find the car's mass. Since we know the weight, we obtain mass m from Equation 4.9:w = mgm = w/g = (19.6 x 10³ N) / (9.80 m/s²) = 2000 kgThen, ΣFₓ = maₓ gives:aₓ = ΣFₓ / m = (-1.50 x 10⁴ N) / 2000 kg = -7.5 m/s²**Figure 4.19**This is a diagram for this problem showing forces acting on the car: - The weight w = 19.6 x 10³N acts downward.- The normal force n = 19.6 x 10³N acts upward.- The force Fₓ = -1.50 x 10⁴N acts to the left (x-direction).**Reflect**This acceleration can be written as -0.77g. Note that 0.77 is also the ratio 1.50 x 10⁴ N to 1.96 x 10⁴ N. An acceleration of this magnitude is possible only on a dry paved road. Don’t expect it over icy roads!**Practical Problem**Passenger comfort requires that the acceleration to stop the car if its speed is initially 30 m/s (roughly 60 mi/h). What distance is required? Answers: 45.9 m, 1960 N.---

Name: The second photo has the background information needed to solve the pr
Uploaded: 2024-03-05T11:56:34.000Z
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Description: The second photo has the background information needed to solve the practice problem. **Practice Problem 4.4:**If passenger comfort requires that the acceleration should be no greater in magnitude than 0.10g, what distance is required to stop the car

Given : Acceleration (a) = 0.10g a = (0.10 × 9.8) m/s2 a = 0.98 m/s2 Velocity (v) = 30 m/s Mass of the car (m) = 2000 kg (given in example)From third equation of motion, we have v2 - u2 = 2ad (30 m/s)2 - 0 = 2 × 0.98 m/s2 × d 900 m2/s2 = 1.96 m/s2 × d d = 900/1.96 m d = 459.18 m d = 459 m Thus, the distance required to stop the car is 459 m.We know that, Force = mass × acceleration Force = 2000 kg × 0.98 m/s2 Force = 1960 N