If Thomas jumps in the air and his position is modeled by the equation y = 3x2 - 12x, how long is Thomas in the air?

Question

If Thomas jumps in the air and his position is modeled by the equation y = 3x2 - 12x, how long is Thomas in the air?

Answer 1

If the problem asks for a time, It seems like the equation should be a function of t (for time) n lieu of x. However, if you assume that x is the time variable, the vertical distance he jumps should be positive and this equation has an incorrect sign. You can easily see this if you substitute x = 1 second:

y = 3(1)² - 12(1) = -9 (which seemingly signifies a negative height)

Nonetheless, ignoring the sign and solving for a height of zero gives:

3x² - 12x = 0

3x(x-12) = 0

So either of the factors must be zero.

3x = 0

x = 0

x - 12 = 0

x = 12

So, Thomas is on the ground at x = 0 seconds and x = 12 seconds and is in the air for the difference, or 12 seconds (which seems quite impossible unless he's on the moon!)

Answer 2

I think you copied the equation incorrectly.