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Cheat answer: velocity is a vector, and can be negative, while KE is a scalar and has to be positive. Therefore you have to square v to get rid of the minus sign. Why not take the absolute value? Nature hates those, probably because the derivative is undefined at 0. So squaring it is.
Cheat answer: velocity is a vector, and can be negative, while KE is a scalar and has to be positive. Therefore you have to square v to get rid of the minus sign. Why not take the absolute value? Nature hates those, probably because the derivative is undefined at 0. So squaring it is.
why not raise to any other even power ?
For the purpose of inverting a negative vector, you can think of squaring as rotating the vector around the unit circle, 180 degrees, to make it positive. Higher order powers just keep rotating that vector back and forth- from this perspective the other even powers are the same transformation. Obviously with the magnitude being different.
Cheat answer: velocity is a vector, and can be negative, while KE is a scalar and has to be positive. Therefore you have to square v to get rid of the minus sign. Why not take the absolute value? Nature hates those, probably because the derivative is undefined at 0. So squaring it is.
Energy is actually not a conserved quantity in Galilean relativity.
Cheat answer: velocity is a vector, and can be negative, while KE is a scalar and has to be positive. Therefore you have to square v to get rid of the minus sign. Why not take the absolute value? Nature hates those, probably because the derivative is undefined at 0. So squaring it is.
Cheat answer: velocity is a vector, and can be negative, while KE is a scalar and has to be positive. Therefore you have to square v to get rid of the minus sign. Why not take the absolute value? Nature hates those, probably because the derivative is undefined at 0. So squaring it is.
That doesn’t answer the title question of why it’s quadratic wrt speed.
Earlier quoted context omitted.
> The blue car, using ½mv², shed (~70²=) 4900 units of energy (we'll hand wave away the constants). So the red car, which had (100²=) 10000 units of kinetic energy to start, also shed 4900 units, which means it had 5100 units of energy when it collided, and so was going (√5100~) 71 But if the cars produce downforce this is no longer true because you brake harder (more friction available) at higher speeds! This is how…
But what if the cars are spherical cows?
Earlier quoted context omitted.
But what if the cars are spherical cows?
I'm sorry to inform you that those cows are going to have a hard time braking on that frictionless surface.
Fun little anecdote: A blue care is travelling along at 70 units, and a red car (exact same make and model) is catching up to it going 100. When they're both right beside each other a bend in the road reveals an obstacle blocking both lanes, so both cars brake at the same intensity and deceleration. The blue care stops right before the obstacle. Since the red car was going at a faster speed, and braked at the same ra…
heh, thats a fun little experiment.