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.

Given a snail is traveling along a straight path. The snail’s velocity can be modeled by [tex]v(t)=1.4ln(1+t^2)[/tex] inches per minute for 0 ≤ t ≤ 15 minutes.

Answer:

B=22.35 Inches per minutes

Step-by-step explanation:

If the snail's velocity is [tex]v(t)=1.4ln(1+t^2)[/tex] per minute, its displacement for 0 ≤ t ≤ 15 minutes is given by the integral:

[tex]\int v(t) dt=\int (1.4ln(1+t^2))dt=76.04307[/tex]

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

The velocity of the ant therefore, twill be:

[tex]\int 2 dt=2t+K, $where K is a constant of integration$[/tex]

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

[tex]\int_{12}^{15}(2t+K) dt=81+3K[/tex]

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