Question

the braking distance (d(v)) (in meters) for a certain car moving at velocity (v) (in meters/second) is given by (d(v)=\frac{v^{2}}{38}). the cars velocity (b(t)) (in meters/second) (t) seconds after starting is given by (b(t) = 5t). write a formula for the braking distance (s(t)) (in meters) after (t) seconds. it is not necessary to simplify. (s(t)=square)

Explanation:

Step1: Identify the relationship

We know that $S(t)=D(B(t))$ since we want to find the braking - distance as a function of time.

Step2: Substitute $v = B(t)$ into $D(v)$

Given $D(v)=\frac{v^{2}}{38}$ and $B(t)=5t$, substitute $v = 5t$ into $D(v)$.
$S(t)=\frac{(5t)^{2}}{38}$

Answer:

$S(t)=\frac{(5t)^{2}}{38}$