An ant arrives at the snail’s starting position at time t=12 minutes and follows the snail’s path. During the interval 12≤t≤15 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 t=15 minutes. The ant’s velocity at time t=12 is B inches per minute. Find the value of B.

B=22.348 Inches per minutes

Step-by-step explanation:

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

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

The ant travels with a constant acceleration of 2 Inches per minute.

Therefore, the velocity of the ant will be:

inches per minutes, for some constant c.

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+3c=76.04307

3c=76.04307-81

3c=-4.95693

c=-1.65231

The velocity of the ant at t=12 is therefore:

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

B=22.348 Inches per minutes

56

step-by-step explanation:

if you are looking for the unit rate then the answer would be 9 because unit rate is distance divided by time.

step-by-step explanation: