Question

The formula for final velocity of an object starting from zero and undergoing constant acceleration is v = 1/2at^2. what type of variation does this represent?

Answer 1

If the velocity is a function of t^2,
then the acceleration is NOT constant.

The correct equation is V = at ,

V = at
50 = a(5)
a = 10 m/sec^2

Answer 2

perhaps that is supposed to be displacement, not velocity, in which case the formula is correct

Answer 3

in any case, in the formula given, v varies directly with t^2

Answer 4

direct variation of thing 1 and thing 2 can be written

thing 1 = thing 2 * constant

Answer 5

I probably should have proof read this.
Fixed-- V = 1/2 (AT) ^2.
Options to choose from:
*Inverse Variation
*Direct Variation
*Combined Variation
*Joint Variation

Answer 6

Sure it isn't \[V = \frac{1}{2}at^2\]?

Answer 7

\(V\) varies directly with \(t^2\) in my formula. \(\frac{1}{2}a\) is the constant of variation.

Answer 8

2v/a = t^2
t = sqrt(2v/a)

Answer 9

So it would be direct variation?

Answer 10

The formula is incorrect. The kinematics equation for constant acceleration in this case would be:\[\bf \Delta x=v_0t+\frac{1}{2}at^2\]Since the object starts from rest, \(\bf v_0=0\), the equation becomes:\[\bf \Delta x = \frac{1}{2}at^2\]There is no "velocity" involved in this equation and like @sadaholic19 stated, if \(\bf v = \frac{1}{2}at^2\) then acceleration is not constant which contradicts the question.