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A police car traveling a constant 95 km/h is passed by a speeder traveling 140 km/h. Precisely 1.00 s after the speeder passes, the policeman steps on the accelerator. If the police car's acceleration is 2.00 m/s2, how much time passes after the speeder passes before the police car overtakes the speeder (assumed moving at constant speed)?
I converted 95 km/h to 40 m/s, and 95 to 26 m/s, then I use the equation
x = x + v*t + 1/2 * a *t^2
40 m/s * t + 40 m = 0 + 26 m/s * t +1/2*2.0*t^2
t=15, but that's wrong....any ideas?
I converted 95 km/h to 40 m/s, and 95 to 26 m/s, then I use the equation
x = x + v*t + 1/2 * a *t^2
40 m/s * t + 40 m = 0 + 26 m/s * t +1/2*2.0*t^2
t=15, but that's wrong....any ideas?