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)=7 t ).
write a formula for the braking distance ( s(t) ) (in meters) after ( t ) seconds.
it is not necessary to simplify.
( s(t)= )

Explanation:

Step1: Substitute \(v = B(t)\) into \(D(v)\)

Since \(S(t)=D(B(t))\) and \(B(t) = 7t\), \(v\) in \(D(v)=\frac{v^{2}}{38}\) is replaced with \(7t\).
So \(S(t)=\frac{(B(t))^{2}}{38}\)

Step2: Write the final formula

Substituting \(B(t) = 7t\) into the above - formula, we get \(S(t)=\frac{(7t)^{2}}{38}\)

Answer:

\(S(t)=\frac{(7t)^{2}}{38}\)