A car is moving 28.5 m/s at some initial position (xi = 0 m). It continues moving this velocity for another 8.55 seconds, but then the driver puts on her brakes. While braking, she effectively has a constant acceleration of -4.72 m/s2. What will be her final position once she comes to a stop?

Sorry my friend i don't know the answer. i am just in 6th

If one were to put the y value into the equation, they could find out what x is for a certain value of y.  putting in the value of y (1), we get the equation: 1=4x-12=4xx=0.5this means that the value of x when why is 1 is x=0.5

The answer is periosteum

The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: where m is the mass of the arrow and v its velocity.  all the other options, instead, do not increase the kinetic energy of the system: -  pulling farther back on the string : this will just increase the elastic potential energy of the system-  holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)-  decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)