How Far Will a Car Travel Before Stopping Without ABS at 115.2 km/h?

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Homework Help Overview

The problem involves a car with a mass of 1000 kg traveling at a speed of 115.2 km/h, which needs to stop after a deer jumps into the road 50 meters ahead. The braking relies solely on the frictional force of 4000 N, and the discussion centers around calculating the stopping distance without the influence of air resistance.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the stopping distance using kinematics and questions their acceleration calculation, which they found to be -5.8 m/s². They express confusion over their results and seek clarification on their approach to the forces acting on the car.

Discussion Status

Participants are engaged in clarifying the calculations and assumptions made by the original poster. One participant requests to see the calculation steps for better assistance, while another confirms the original poster's understanding of the forces involved in stopping the car.

Contextual Notes

There is a mention of a previous thread that may relate to similar calculations, indicating that this is a recurring topic of discussion. The original poster expresses confusion over their acceleration calculation, suggesting a need for further exploration of the concepts involved.

David Donald
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Homework Statement



While traveling on the highway with your 1000kg car, at 115.2 km/h, where you’re ABS (automatic

braking system) is disabled. This means braking is relying solely on the friction of your tires with the road when they stop spinning. A dear jumps into the road 50 meters in front of you. If the frictional force created by you slamming on your brakes is 4000N. What will your final stopping distance be? Will you

hit the dear? Assume no air resistance.

Homework Equations



Kinematics?

The Attempt at a Solution


Sum of Forces in The X direction
(Force O' Car) - (Force O' Friction) = -ma

I solved for acceleration and got -5.8 m/s^2
plugging these into the kinematics equation I got a time... 5.52 seconds
plugging that into the Xf = Xo + Vox t + 1/2 a t^2 I got a distance which is wrong

what gives? what am i doing wrong?
 
Physics news on Phys.org
Your acceleration is wrong. Please show your calculation steps so we can provide more targeted assistance.
 
When someone tells you, "Be a dear," they don't mean for you to drop down on all fours while wearing a hat rack on your head. :rolleyes:

"Deer" is the animal which jumps out in front of the car. :wink:
 
Ok... So I am not able to get to the same acceleartion I was getting before so now I'm really confused

Since the only thing acting on the car when its stopping is the breaking force would it be

Sum Fx = -Fbrakes = ma ?
 
That's right.
 

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