Question

The distance a spring will stretch Waries directly with how much weight is attached to the spring. If a springstretches 5 inches with 95 pounds attached, now far will it stretchwith 55 pounds attached? Round to themearest tenth of an inch.

Answer 1

From the first sentence of the exercise, we know that this is a question about direct variation. Then, we can write the following equation:

$\begin{gathered} d=k\cdot w \\ \text{ Where} \\ d\text{ is the distance} \\ k\text{ is the constant of variation and} \\ w\text{ is the weight} \end{gathered}$

We can find the value of k replacing the known values in the above equation:

$\begin{gathered} d=5 \\ w=95 \\ d=k\cdot w \\ 5=k\cdot95 \\ 5=95k \\ \text{ Divide by 95 from both sides of the equation} \\ \frac{5}{95}=\frac{95k}{95} \\ \frac{1\cdot5}{19\cdot5}=k \\ \frac{1}{19}=k \end{gathered}$

Now, we can find the new value of d replacing w = 55 and the found value of k in the initial equation:

$\begin{gathered} k=\frac{1}{19} \\ w=55 \\ d=k\cdot w \\ d=\frac{1}{19}\cdot55 \\ d=\frac{55}{99} \\ \text{ Or aproximately} \\ \end{gathered}$