QUESTION 10
A motorist traveling north at $32.0 \mathrm{~m} / \mathrm{s}$ spots a police car $110.0 \mathrm{~m}$ ahead traveling south at $27.0 \mathrm{~m} / \mathrm{s}$. The motorist decelerates at 2.0 $\mathrm{m} / \mathrm{s}^{2}$ in an attempt to avoid a ticket. How fast is the motorist traveling when they pass the police car?

Step 1 ：First, calculate the relative speed of the motorist and the police car, which is the sum of their speeds. This is given by \(32.0 \, m/s + 27.0 \, m/s = 59.0 \, m/s\).

Step 2 ：Next, calculate the time it takes for them to meet. This is given by the distance divided by the relative speed, which is \(110.0 \, m / 59.0 \, m/s = 1.86 \, s\).

Step 3 ：Then, calculate the change in speed of the motorist due to deceleration. This is given by the deceleration multiplied by the time, which is \(2.0 \, m/s^2 * 1.86 \, s = 3.72 \, m/s\).

Step 4 ：Finally, calculate the speed of the motorist after deceleration. This is given by the initial speed minus the change in speed, which is \(32.0 \, m/s - 3.72 \, m/s = 28.28 \, m/s\).

Step 5 ：So, the motorist is traveling at \(\boxed{28.28 \, m/s}\) when they pass the police car.