Let's denote the initial velocities of car P and car Q as v_Pi and v_Qi, respectively. Also, denote the accelerations of car P and car Q as a_P and a_Q, respectively. Moreover, the displacement covered by car P is represented as Δx_P and that of car Q as Δx_Q.

In order to tie in with car Q, car P should take the same time to reach the cabin as car Q i.e., t_P = t_Q = t where, t_P and t_Q represent the time for car P and car Q, respectively.

Using 2nd equation of motion to find the time taken by car Q;

Δx_Q = v_Qit + (1/2)a_Qt²

76 m = [(55 m/s)t] + [(½)(-3.2 m/s²)t²]

Rearranging this equation gives a quadratic equation with variable t as follow;

1.6t² - 55t + 76 = 0

Solving this quadratic equation, we get;

t = 32.93 s or t = 1.442 s

We will take t = 1.442 s because t = 32.93 s represents the time taken by car Q as it keeps decelerating and then returns to the cabin. Hence, car Q takes 1.442 s to reach the cabin as it covers the distance of 76 m. Thus, the time for car P to tie in with car Q is t = 1.442 s.

Now we that we know the value of time required for car P, we can find its acceleration using the following equation of motion;

Δx_P = v_Pit + (1/2)a_Pt²

98 m = [(45 m/s)(1.442 s)] + [(½)(a_P)(1.442 s)²]

Solving a_P for gives;

a_P = 31.85 m/s²≈ 32 m/s²

A police officer at rest on a sidewalk is passed by a thief running at a steady speed of 3.50 m/s. Just as the thief exceeds him, the police officer accelerates at a rate of 1.00 m/s2. Determine the distance the police officer travels before catching the thief.

A police officer at rest on a sidewalk is passed by a thief running at a steady speed of 3.50 m/s. Just as the thief exceeds him, the police officer accelerates at a rate of 1.00 m/s². Determine the distance the police officer travels before catching the thief.