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THE LOCK NUT SYSTEMThe lock-nut system is by far the most accurate braking system that can be used. The lock-nut braking system uses a wing-nut that travels along a threaded axle. As the threaded axle turns a wing nut travels along the threaded axle until it lock tight against the frame of the vehicle, or the axle, causing the vehicle to come to an instant stop. Setting the braking system is simple and easy: just place the vehicle at the finish line with the wing nut in the lock position, roll the vehicle back to the start line and the wing nut is set to the exact starting position. Now you are ready to go. The wing-nut braking is very accurate and consistent but the best part it that it can be set for almost any distance by following the same steps as described.

Setting the lock-nut braking system is simple and easy, just place the vehicle at the finish line with the wing nut in the locked position, roll the mousetrap vehicle back to the start line, and this sets the wing nut to the exact starting position needed. This easy to set braking system can be re positioned for any length course by repeating the described procedure.

Adding traction to a wheel will give the wheel more grip and greater stopping potential. One way to increase the traction of a wheel on smooth surface is to add a rubber traction tread that has more grip than the original wheel alone. Various brands of rubber may have more grip than another brands of rubber so it is always best to experiment. Rubber bands can be stretches and placed directly over a wheel or the middle section of a balloon can be cut and then also stretch around a wheel to help increase the traction

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Q: What is a braking system not involving the end of any string that will stop a mousetrap car at a certain distance?
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What happen to the braking distance if the speed is doubled and why?

Hi there! Assuming that the deceleration (or negative acceleration, if you will) is constant and the same in both cases, you can use a special kinematic formula to solve the problem. The formula follows: (final velocity)^2 = (initial velocity)^2 + [ 2 * (deceleration) * (braking distance) ] Rearranged to our needs the formula reads: braking distance = [1/2] * -(initial velocity)^2 / (deceleration) * this equation assumes that the final velocity is zero If the initial speed were doubled then the general formula would read: braking distance = 2 * -(initial velocity)^2 / (deceleration) NOTICE that the two equations are the exact same except for the leading coefficients. 1/2 is assocaited with the braking distance of the normal velocity while 2 is assocated with the breaking distance of the doubled velocity. Since 2 is four times larger than 1/2, this leads us to the conclusion that the breaking distance for an object traveling at double a certain velocity would be 4x greater than the breaking distance of the object moving at the "regular" velocity.


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