The distance, d, through which a stone falls from rest varies directly as the square of the time, t, taken. If…

Analyze the question: The distance d varies directly as the square of time t, so d = kt².

Concept: In direct variation, d = kt², where k is a constant. Find k from the first set of data, then use it to find the new distance.

Solution:

Step 1: Find k using d = 45 when t = 3:

45 = k(3²) = 9k

k = 45 ÷ 9 = 5

Step 2: Use d = 5t² with t = 6:

d = 5(6²) = 5 × 36 = 180 cm

Check with a ratio (a quicker method): Time doubles (3 → 6), so t² becomes 4 times larger. Distance is therefore 4 × 45 = 180 cm.

Why the other options are wrong:

• A (90 cm) comes from doubling the distance, which would be direct variation with t, not t².
• B (135 cm) comes from tripling the distance, which is not what happens here.
• D (225 cm) has no correct basis.

Tip: When y varies as xⁿ, a change in x by a factor of m changes y by a factor of mⁿ. Here, time × 2 means distance × 2² = × 4. This saves time in the exam, as you can skip finding k.