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**Homework Statement**

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}

## Homework Equations

x = x

_{0}+ v

_{0}t + 0.5 * a * t

^{2}

## 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...