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Sagot :
Assuming that the force of friction was the only force on the horizontal plane we have, using Newton's second law, that:
[tex]-F_f=ma[/tex]The minus sign indicates that the force is going againts the motion.
Now, the force of friction is given by:
[tex]F_f=\mu N[/tex]where mu is the coeffient of friction and N is the normal force. In this case the normal force is equal to the weight of the car, then our equation of motion is:
[tex]\begin{gathered} \\ -\mu mg=ma \end{gathered}[/tex]Solving for a we have that:
[tex]\begin{gathered} a=-\mu g \\ a=-(0.486)(9.8) \\ a=-4.7628 \end{gathered}[/tex]Therefore the acceleration is -4.7628 meters per second per second.
Now that we know the acceleration we can use the formula:
[tex]v^2_f-v^2_0=2a(x-x_0)_{}[/tex]To find how fast the car was going. In this case the final velocity is zero and the change in position is 55 meters, plugging the values we have that:
[tex]\begin{gathered} 0^2-v^2_0=2(-4.7628)(55) \\ v^2_0=523.908 \\ v_0=\sqrt[]{523.908} \\ v_0=22.9 \end{gathered}[/tex]Therefore the initial velocity of the car was 22.9 meters per second.
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