Certainly! Let's solve this step-by-step.
1. Start with the given formulas:
- The braking distance [tex]\( D(v) \)[/tex] is given by:
[tex]\[
D(v) = \frac{v^2}{26}
\][/tex]
- The car's velocity [tex]\( B(t) \)[/tex] at time [tex]\( t \)[/tex] seconds is given by:
[tex]\[
B(t) = 3t
\][/tex]
2. We need to find the braking distance [tex]\( S(t) \)[/tex] as a function of time [tex]\( t \)[/tex]. To do this, we will substitute the velocity function [tex]\( B(t) \)[/tex] into the braking distance function [tex]\( D(v) \)[/tex].
3. Substitute [tex]\( B(t) \)[/tex] into [tex]\( D(v) \)[/tex]:
- Since [tex]\( v = B(t) \)[/tex], we can rewrite [tex]\( v \)[/tex] in the braking distance formula:
[tex]\[
v = 3t
\][/tex]
- Now, substitute [tex]\( v = 3t \)[/tex] into the braking distance formula [tex]\( D(v) \)[/tex]:
[tex]\[
D(3t) = \frac{(3t)^2}{26}
\][/tex]
4. Simplify the expression:
[tex]\[
D(3t) = \frac{(3t)^2}{26}
= \frac{9t^2}{26}
\][/tex]
5. Therefore, the braking distance [tex]\( S(t) \)[/tex] as a function of time [tex]\( t \)[/tex] is:
[tex]\[
S(t) = \frac{9t^2}{26}
\][/tex]
So the answer is:
[tex]\[
S(t) = \frac{9t^2}{26}
\][/tex]
