Sensors are 90 meters apart in a school zone with a 9 m/s speed limit. If Vehicle A takes 8 s and Vehicle B takes 12 s to travel 90 m, what is each vehicle’s speed in m/s, and which one exceeded the limit and by how much?
Vehicle A’s speed is $v=\frac{90}{8}=11.25\,\text{m/s}$ and Vehicle B’s speed is $v=\frac{90}{12}=7.5\,\text{m/s}$. The safe limit is $9\,\text{m/s}$, so Vehicle A exceeded the limit by $11.25-9=2.25\,\text{m/s}$, while Vehicle B did not exceed the limit.
What you’re doing here
You’re given a distance between sensors and the time each vehicle takes to travel that distance. Speed is found by dividing distance by time, then you compare each speed to the posted limit.
Compute each speed from distance and time
Use $$v=\frac{d}{t}$$ where $d=90\,\text{m}$.
Vehicle A: $$v_A=\frac{90}{8}=11.25\,\text{m/s}$$
Vehicle B: $$v_B=\frac{90}{12}=7.5\,\text{m/s}$$
Compare to the 9 m/s safety limit
Limit: $9\,\text{m/s}$.
- Vehicle A: $11.25 > 9$, so it is speeding. $$11.25-9=2.25\,\text{m/s} \text{ over the limit}$$
- Vehicle B: $7.5 < 9$, so it is not speeding.
Quick reasonableness check
Covering the same 90 m in less time means a higher speed. Since 8 s is less than 12 s, Vehicle A should be faster, which matches the calculations.
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