An ant arrives at the snail’s starting position at time minutes and follows the snail’s path. During the interval minutes, the ant travels in the same direction as the snail with a constant acceleration of 2 inches per minute per minute. The ant catches up to the snail at time minutes. The ant’s velocity at time is B inches per minute. Find the value of B.

QUESTION BEGINNING

Given a snail is traveling along a straight path. The snail’s velocity can be modeled by inches per minute for 0 ≤ t ≤ 15 minutes.

B=22.35 Inches per minutes

Step-by-step explanation:

If the snail's velocity is per minute, its displacement for             0 ≤ t ≤ 15 minutes is given by the integral:

The constant acceleration of the ant is 2 Inches per minute.

The velocity of the ant therefore, twill be:

For the interval, 12≤t≤15, the displacement of the ant is:

Since the snails displacement and that of the ant are equal in 12≤t≤15.

81+3K=76.04307

3K=76.04307-81

3K=-4.95693

K=-1.65231

At t=12, the velocity of the ant  is therefore:

2t+K=2(12)-1.65231=22.348 Inches per minutes

B=22.348 Inches per minutes

c - they will have the same distribution

step-by-step explanation:

this is the only one that could make sense. : )