Final answer:
To determine the police car's acceleration, kinematic equations are used, equating the displacement of both Bob and the police car after meeting at 81 m. Bob's constant velocity results in a travel time of 13.5 s, which is used to calculate the police car's acceleration to be approximately 0.89 m/s².
Step-by-step explanation:
To find out the police car's acceleration, given that Bob is going at a constant velocity of 6 m/s and they meet at 81 m, we must apply kinematic equations. We have two objects: Bob moving at a constant speed, and the police car starting from rest with a constant acceleration. The equation for displacement for Bob, who is not accelerating, is:
x = vt
For the police car, which starts from rest and accelerates, the equation for displacement is:
x = ½ at²
Since they meet after traveling the same distance (81 m), we can set these two equations equal to each other:
6t = ½ a t²
By dividing both sides by t and multiplying by 2, we get:
12 = at
Now, since Bob travels 81 m at a constant speed of 6 m/s, the time (t) it takes him to travel that distance is:
t = 81 m / 6 m/s = 13.5 s
Substituting this time into our equation for a gives:
a = 12 / 13.5 s
Hence, the acceleration of the police car is:
a ≈ 0.89 m/s²