# Physics question help?

## An unmarked police car travelling at a constant 80.0 km/h is passed by a speeder travelling at 100.0 km/h. Precisely 1.0 s after the speeder passes, the police steps on the accelerator. If the police car’s acceleration is 2.0 m/s^2, how much time passes before the police car overtakes the speeder (assumed moving at a constant speed)?

Mar 27, 2017

The velocity of police car

${v}_{p} = 80 k m \text{/"h=(80xx10^3)/3600m"/"s=200/9m"/} s$

The velocity of the speeder

${v}_{s} = 100 k m \text{/"h=(100xx10^3)/3600m"/"s=250/9m"/} s$

1.0 s after the speeder passes the police car the later starts accelerating @ $2 m \text{/} {s}^{2}$.

Within this 1.0 s the speeder goes $\left(\frac{250}{9} - \frac{200}{9}\right) m = \frac{50}{9} m$ ahead of the police car. Let the police car reaches the speeder again after $t$ sec, it starts accelerating.

The distance covered by the police car during t sec after it accelerting @$a = 2 m \text{/} {s}^{2}$

${S}_{p} = {v}_{p} \times t + \frac{1}{2} a {t}^{2} = \frac{200}{9} t + \frac{1}{2} \cdot 2 \cdot {t}^{2}$

$= \frac{200}{9} t + {t}^{2}$

The distance covered by the speeder during the same t sec will be

${S}_{s} = \frac{250}{9} t$

By the condition of the problem

${S}_{p} - {S}_{s} = \frac{50}{9}$

$\implies {t}^{2} + \frac{200}{9} t - \frac{250}{9} t = \frac{50}{9}$

$\implies 9 {t}^{2} - 50 t - 50 = 0$

$\implies t = \frac{50 + \sqrt{{50}^{2} + 4 \cdot 9 \cdot 50}}{18} s = 6.42 s$

So if T represets the total time passed before the police car overtakes the speeder then

$T = 1 \sec + 6.42 s = 7.42 s$