(adsbygoogle = window.adsbygoogle || []).push({}); The problem statement, all variables and given/known data

An unmarked police car traveling a constant 95km/h is passed by a speeder traveling 140 km/h. Precisely 1.00s after the speeder passes, the policeman steps on the accelerator; if the police car's acceleration is 2.00m/s^{2}, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)?

v_{0p}= 95 km/h = 26.389 m/s

v_{0s}= 140 km/h = 38.889 m/s

a_{p}= 2.00 m/s

t_{1}= 1.00 s

v_{s}= v_{0s}

2. Relevant equations

x = x_{0}+ v_{0}t + 0.5 * a * t^{2}

3. The attempt at a solution

Let t_{2}= the time it takes them to meet after the initial 1.00s.

After 1.00s, the x_{p}= 26.389m and x_{s}= 38.889m. Taking these two quantities as our initial positions, x_{0p}and x_{0s}, we have:

x_{p}= x_{0p}+ v_{0p}t_{2}+ 0.5 * a_{p}* t_{2}^{2}

x_{s}= x_{0s}+ v_{0s}t_{2}

Setting these two equal, I get a quadratic equation and find t_{2}= 13.43 s or -0.9307 s. The total time would then be 14.43s...

But wait. Shouldn't the negative answer for t_{2}be -1.00s?? Because 1.00s in the past, the cars met... So I must be doing something wrong...

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# Homework Help: Police car and speeder when do they meet?

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