Let's solve this problem step-by-step to identify where Nick might have gone wrong.
1. According to the problem, [tex]\( s \)[/tex] varies directly as the square of [tex]\( t \)[/tex]. This means we can write the equation as:
[tex]\[ s = k t^2 \][/tex]
2. We know that [tex]\( s = 4 \)[/tex] when [tex]\( t = 12 \)[/tex]. Let's use these values to find the constant of variation [tex]\( k \)[/tex]:
[tex]\[
4 = k \cdot 12^2
\][/tex]
Rewriting the equation:
[tex]\[
4 = k \cdot 144
\][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[
k = \frac{4}{144} = \frac{1}{36}
\][/tex]
So, we have determined that the constant [tex]\( k \)[/tex] is [tex]\( \frac{1}{36} \)[/tex].
3. Now, we need to find the value of [tex]\( t \)[/tex] when [tex]\( s = 48 \)[/tex]. We use the equation [tex]\( s = k t^2 \)[/tex] and substitute the known values:
[tex]\[
48 = \frac{1}{36} t^2
\][/tex]
Multiplying both sides by 36 to isolate [tex]\( t^2 \)[/tex]:
[tex]\[
48 \times 36 = t^2
\][/tex]
[tex]\[
1728 = t^2
\][/tex]
Taking the square root of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[
t = \sqrt{1728} \approx 41.57
\][/tex]
Let's compare Nick's work:
Nick wrote:
[tex]\[
12 = k \cdot 4^2
\][/tex]
Here, he should have written:
[tex]\[
4 = k \cdot 12^2
\][/tex]
From which he gets:
[tex]\[
12 = 16k
\][/tex]
[tex]\[
k = 0.75
\][/tex]
His error is in step 2 when calculating the constant of variation; he incorrectly used [tex]\( t \)[/tex] and [tex]\( s \)[/tex] values. Therefore, the first error in Nick's work is:
C. He substituted incorrectly when calculating the constant of variation.
