Question

The distance versus time plot for a particular object shows a quadratic relationship. Which column of distance data is possible for this situation? Time (s) A. Distance (m) B. Distance (m) C. Distance (m) D. Distance (m) E. Distance (m) 0 0 2.00 9.00 ‡ ‡ 1 1.00 4.00 18.00 1.00 1.00 2 4.00 6.00 27.00 0.50 0.25 3 9.00 8.00 36.00 0.33 0.11 4 16.00 10.00 45.00 0.25 0.06 5 25.00 12.00 54.00 0.20 0.04 6 36.00 14.00 63.00

Answer 1

I think the answer will have to be d 

Answer 2

Answer:

COLUMN A

Explanation:

The distance versus time plot for a particular object shows a quadratic relationship. Which column of distance data is possible for this situation? Time (s) A. Distance (m) B. Distance (m) C. Distance (m) D. Distance (m) E. Distance (m) 0 0 2.00 9.00 ‡ ‡ 1 1.00 4.00 18.00 1.00 1.00 2 4.00 6.00 27.00 0.50 0.25 3 9.00 8.00 36.00 0.33 0.11 4 16.00 10.00 45.00 0.25 0.06 5 25.00 12.00 54.00 0.20 0.04 6 36.00 14.00 63.00

A quadratic relation mean one variable is proportional to the square of the other variable.

for this problem, distance and time will be proportional to the following equation

d = kt²

or

t = kd²

where k is a (constant).

when y look  at  column A closely .  The numbers are all perfect squares: 1, 4, 9, 16...

This means d = kt²  with k=1:

at t = 0, d = 1 x 0² = 0

at  t = 1, d = 1 x 1² = 1

at  t = 2, d = 1 x 2² = 4

at   t = 3, d = 1 x 3² = 9