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RV-14-001 · Inverse-square model and a changed condition

A sensor reading S is inversely proportional to the square of its distance d from a source. When d = 10 m, S = 29.

(a) Find S when d = 17 m. Give your answer to 3 decimal places.

(b) The distance from the source is increased by 50%. Find the percentage decrease in the reading, to 2 decimal places.

Part (a) Inverse-Square Proportion & Percentage Change Booklet p. 23
Method
1) Replace ∝ by a constant k
2) Substitute the known pair to find k
3) Use the formula with the new distance
Worked Example (RV-14-001 · Part (a))
Find S when d = 17 m. Give your answer to 3 decimal places.
1)
S=kd2S=\frac{k}{d^2}
2)
Substitute the known pair: 29=k102⇒k=290029=\frac{k}{10^2}\Rightarrow k=2900.
3)
S=2900172=10.035S=\frac{2900}{17^2}=10.035
Ans = 10.035
Part (b) Inverse-Square Proportion & Percentage Change Booklet p. 23 & p. 52
Method
1) Find the distance multiplier
2) Use its reciprocal square for the reading multiplier
3) Convert the decrease to a percentage
Worked Example (RV-14-001 · Part (b))
The distance from the source is increased by 50%. Find the percentage decrease in the reading, to 2 decimal places.
1)
Distance multiplier = 1.51.5. The reading multiplier is 11.52\frac{1}{1.5^2}.
2)
Decrease = (1−11.52)×100=55.56%\left(1-\frac{1}{1.5^2}\right)\times100=55.56\%.
Ans = 55.56%