I have mass and velocity, how do I get accel?

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To find the acceleration of a car with a mass of 1750 kg traveling at 110 km/h on a slippery road, the friction force is calculated as 25% of the car's weight. Using Newton's law, the acceleration is determined to be -1/4g m/s², which equates to approximately -2.45 m/s². To calculate the stopping distance, the equation Vf - Vi = 2*a*delta x is applied, with the final velocity (Vf) set to zero. The discussion highlights the importance of understanding the role of friction as the sole force acting on the vehicle during deceleration. This approach clarifies the relationship between mass, friction, and acceleration in physics problems.
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Homework Statement


The driver of a 1750 kg car traveling on a horizontal road at 110 km/h suddenly applies the brakes. Due to a slippery pavement, the friction of the road on the tires of the car, which is what slows down the car, is 25% of the weight of the car.



Homework Equations



(a) What is the acceleration of the car?

(b) How many meters does it travel before
stopping under these conditions?


The Attempt at a Solution


I keep getting -2.7m/s^2 and i cannot figure out the correct way to do it.
 
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knight4life said:

The Attempt at a Solution


I keep getting -2.7m/s^2 and i cannot figure out the correct way to do it.
How did you get that value? (Hint: Use Newton's law.)
 
Doc Al said:
How did you get that value? (Hint: Use Newton's law.)

I don't know, I cannot seem to get that value again.
 
For part a, you do not even need the velocity of the truck. If you use Newton's law like Doc Al says, you will have -F(friction)=ma --> F(friction)=-ma, which makes sense because the truck is decelerating. You know the force of friction is (1/4)mg so a should equal (-1/4)g m/s^2. Then to find the distance traveled, use the equation Vf-Vi=2*a*delta x, where Vf is final velocity and Vi is initial velocity. In this case, your final velocity will be zero.
 
w3390 said:
For part a, you do not even need the velocity of the truck. If you use Newton's law like Doc Al says, you will have -F(friction)=ma --> F(friction)=-ma, which makes sense because the truck is decelerating. You know the force of friction is (1/4)mg so a should equal (-1/4)g m/s^2.

Thank you very much. I was trying to figure out using kinematics, and getting very angry. I also was assuming that she wanted the acceleration of the vehicle, not that of the friction force.
 
That is the acceleration of the vehicle. The friction force is used because it is the only force acting on the truck.
 
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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