A police officer spots a speeding car traveling at 20 m/s (approximately 44 mph)as it passes.The officer immediately starts to drive,accelerating from rest at a rate of 5 m/s².How far will the officer have to drive before catching up to the speeding car?

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Quizplus · Accepted Answer

To solve this, we use the equations of motion. Let $t$ be the time in seconds it takes for the officer to catch up to the speeding car. The distance covered by the speeding car in this time is $20t$ meters (since speed = distance/time, and the car's speed is 20 m/s). The officer starts from rest and accelerates at $5 \, \text{m/s}^2$, so the distance covered by the officer can be found using the equation $d = \frac{1}{2}at^2$, where $a$ is the acceleration and $t$ is the time. Substituting $a = 5 \, \text{m/s}^2$, we get the officer's distance as $\frac{1}{2} \times 5 \times t^2 = \frac{5}{2}t^2$.For the officer to catch up to the car, the distances covered by both must be equal, so we set them equal to each other and solve for $t$:$$20t = \frac{5}{2}t^2$$Multiplying both sides by 2 to clear the fraction:$$40t = 5t^2$$Rearranging:$$5t^2 - 40t = 0$$$$t(5t - 40) = 0$$This gives $t = 0$ (which we ignore since it's the start time) and $t = 8$ seconds as the solutions. Substituting $t = 8$ seconds back into the distance formula for either the car or the officer gives the distance:$$20 \times 8 = 160 \, \text{m}$$Therefore, the officer will have to drive 160 meters to catch up to the speeding car.

A Police Officer Spots a Speeding Car Traveling at 20