Question

Calculus - help with part b only
[see attachment]

Answer 1

I know d/t = v takes place to help find the answer. The faster B is the farther from a until a certain point.

Answer 2

As long as car B drives faster than car A the distance between them will _increase_.
At t = 2.5 the distance between them is constant and then as the speed of car A becomes faster than that of car B the distance will _decrease_.

Answer 3

so would the answer be at 2.5 when it is the greatest?

Answer 4

I think it's a little more than 2.5.

Answer 5

\[Distance = \int\limits_{0}^{t}(V_B -V_A) dt\]
to maximize distance, set derivative equal to 0
\[\frac{d}{dt} \int\limits\limits_{0}^{t}(V_B -V_A) dt = 0\]
\[V_B -V_A = 0\]
\[V_B = V_A\]
so max distance occurs at time when Car A velocity catches up to car B velocity

Answer 6

thank you.