Stopping Distance and kinetic friction

AI Thread Summary
The discussion revolves around calculating the stopping distance of a 1000kg car traveling at 115.2 km/h with a frictional force of 4000N, assuming the ABS is disabled. The initial calculations yielded an acceleration of -5.8 m/s², but this was questioned as incorrect. Participants noted that the acceleration due to the frictional force on the car should be recalculated, as the initial assumption of constant velocity does not apply when brakes are engaged. The need for accurate calculations and understanding of forces in motion is emphasized, particularly in relation to the stopping distance and whether the car would hit the deer. Accurate calculations are crucial for determining the final stopping distance.
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?
 
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Maybe look at the momentum of the car. Units will be N sec. What would be a reasonable way to find seconds to stop under friction force in N?
I get something greater than 6 seconds.
 
David Donald said:

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?
We have no idea; you didn't provide your complete calculations.

BTW, "you're" = "you are" and shouldn't be used to mean "your".
 
By the way...what is the acceleration caused by 4000N on a 1000kg car? If the car is starting at a constant velocity, that means acceleration of the car is zero...so what is the acceleration in the system? It is not -5.8 m/sec^2.
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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