Find for how much time, the car with stop and how many meter

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SUMMARY

The discussion centers on calculating the stopping time and distance for a car with a mass of 1000 kg traveling at 50 km/h, using a braking force of 210 N per wheel. The initial calculation yielded a stopping time of 66.14 seconds, which was incorrect. The correct approach involves recognizing that the total braking force from all four wheels is 2800 N, leading to a deceleration of 2.8 m/s² and a stopping time of 4.96 seconds, consistent with the book's solution.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Basic knowledge of kinematics (v = u + at)
  • Familiarity with the concept of torque and its application in braking systems
  • Ability to convert units (e.g., km/h to m/s)
NEXT STEPS
  • Study the principles of torque and its effects on braking systems in vehicles
  • Learn about kinematic equations for uniformly accelerated motion
  • Explore the relationship between force, mass, and acceleration in practical scenarios
  • Investigate the impact of multiple braking systems on overall stopping distance and time
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Students in physics or engineering, automotive engineers, and anyone interested in vehicle dynamics and braking systems.

MaiteB
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Homework Statement


A car with mass 1000kg moves with velocity 50 km/hour. The system of deceleration in a car is based on the principle of "force match" that is applied on each of four wheels with a force of 210 N and the ray of the wheel is 0.3m. Find for how much time, the car with stop and how many meters it has done. I thought this: a=F/m=0.21, a=v/t=> t=66,14 S but in my book solution it is 4.96 S.

Homework Equations


a=F/m, a=v/t

The Attempt at a Solution


I thought this: a=F/m=0.21, a=v/t=> t=66,14 S but in my book solution it is 4.96 S
 
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Hello Maite,

That is a very interesting deceleration principle, but I have never heard of it and can't even find a hint of it anywhere.

But in your working, I see you only use one wheel to brake. Using four wheels would already reduce the braking time to 16.53 s.

And then the ratio of the book answer to your result is exactly 0.3.

Could it be that the braking system exerts a torque of 210 Nm on each of the wheels, which -- with a wheel radius of 0.3 m -- translates into a braking force of 700 N for each of them, so 2800 N in total ?
 

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