Aggressive braking will wear brakes down prematurely. So the answer would be yes.
Wheel hubs or bearings or could be the CV shafts.
Stupid loads would affect a structure!
If you only did the front brakes, the problem may be in the rear brakes.
There are many different forms of arthritis and what would affect some would have no affect on others.
The answer depends on compared to what? Compared to driving at 50 km per hour, the braking (not breaking!) distance would be longer, compared to 200 km per hour it would be longer.
Total stopping distance includes both reaction distance (the distance traveled while perceiving a hazard and reacting to it) and braking distance (the distance traveled once braking has been initiated). It is the sum of these two distances and is the distance required for a vehicle to come to a complete stop.
On dry pavement in the average car it will take 60 ft of thinking about it, & 180 ft of braking for a total of 240 ft. Double the braking distance on wet pavement for a total of 420 ft. On snow it is anyone's guess.
At 20 mph, the average thinking distance is around 20 feet, while the braking distance is approximately 20 feet as well. Therefore, the overall stopping distance for a vehicle traveling at 20 mph would be around 40 feet.
In a word, no.
because there is less friction between the tyre and the road because of the water in between
I am trying to understand your question and interpret it as meaning: How does the reaction time affect the breaking distance of a car at different speeds. The simple answer is that the reaction time "thinking distance" does not change, but the distance a car travels at higher speeds changes during that time does. For example: If you are too close to the car in front of you and they slam on their breaks, if you are both going fast enough, by the time you did your "thinking time" you would be crashing into their rear end.
If you're running long distance in track they would be good for that. But if you're doing short distance runs, but fast, I would recommend spiked running shoes.
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.
Yes, because an object would decrease the lift wind would give, so there would be less distance
a shorter radius would mean a shorter track distance. The smaller the radius - the smaller the circumference.
If we're talking about a standard track that measures 440 yards, then the distance traveled would be about 1.57 miles.