An ant arrives at the snail’s starting position at time minutes and follows the snail’s path. During the interval minutes, the a

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An ant arrives at the snail’s starting position at time minutes and follows the snail’s path. During the interval minutes, the a

nt 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.

Mathematics

1 answer:

babunello [11.8K] · Answer 1 · 2026-02-18T13:25:33+03:00

QUESTION INTRODUCTION

A snail is moving in a straight line. Its speed can be represented by $v(t)=1.4ln(1+t^2)$ inches per minute for the duration of 0 ≤ t ≤ 15 minutes.

Response:

B=22.35 inches per minute

Detailed Explanation:

Assuming the snail moves at $v(t)=1.4ln(1+t^2)$ inches per minute, its displacement from 0 to 15 minutes is determined by the integral:

$\int v(t) dt=\int (1.4ln(1+t^2))dt=76.04307$

The ant's acceleration remains constant at 2 inches per minute squared.

This results in the ant's velocity as:

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

During the period of 12≤t≤15, the ant’s displacement is:

$\int_{12}^{15}(2t+K) dt=81+3K$

Given that the displacement of the snail matches that of the ant during 12≤t≤15.

81+3K=76.04307

3K=76.04307-81

3K=-4.95693

K=-1.65231

At t=12, the ant’s speed becomes:

2t+K=2(12)-1.65231=22.348 inches per minute

B=22.348 inches per minute