A trooper is moving due south along the freeway at a speed of 23 m/s. at time t = 0, a red car passes the trooper. the red car m

Result: The maximum distance reached is 13.9m

Details: The initial velocity of the trooper is 23m/s, while the red car has an initial velocity of 31m/s, both heading south, so this is a one-dimensional situation.

The trooper begins accelerating at t = 0s, with a rate of 2.3m/s².

Next, we determine the position equations for both vehicles by integrating with respect to time (assuming they start from the same initial position, which means no need for integration constants).

For the red car:

v(t) = 31m/s

p(t) = 31m/s*t

For the trooper:

A(t) = 2.3m/s²

V(t) = 2.3m/s²*t + 23m/s

P(t) = 1.15m/s²*t² + 23m/s*t

We want to find the time when the difference in positions, p(t) and P(t), is maximized. This occurs when the trooper's velocity equals that of the red car; thereafter, the trooper's velocity exceeds that of the car, leading to a decrease in the separation distance. Initially, we have:

2.3m/s²*t + 23m/s = 31m/s

2.3m/s²*t = 8m/s

t = (8/2.3)s = 3.48s

The maximum distance can be calculated as:

p(3.48s) - P(3.48s) = (31*3.48 - 1.15*3.48² - 23*3.48)m = 13.9m