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# A car with speed v = "20 ms"^(-1) passes a stopped police car. Right at this point, the car accelerates at "2 ms"^-2 and the police car at 2a. Show that, when the police car catches up to the other car, the distance is d=(4v^2)/a?

Mar 8, 2018

Let the police car catch up with the other car after time $t$ after it starts.
Applicable kinematic equation is

$s = u t + \frac{1}{2} a {t}^{2}$

1. For Police car
$d = 0 \times t + \frac{1}{2} \left(2 a\right) {t}^{2}$
$\implies d = a {t}^{2}$
$\implies t = \sqrt{\frac{d}{a}}$ .....(1)

2. For car

$d = v t + \frac{1}{2} a {t}^{2}$

Insert value of $t$ from (1)

$d = v \sqrt{\frac{d}{a}} + \frac{1}{2} a \frac{d}{a}$
$\implies \frac{d}{2} = v \sqrt{\frac{d}{a}}$

Squaring both sides we get

${\left(\frac{d}{2}\right)}^{2} = {\left(v \sqrt{\frac{d}{a}}\right)}^{2}$
$\implies {d}^{2} = \frac{4 {v}^{2}}{a} d$
$\implies {d}^{2} - \frac{4 {v}^{2}}{a} d = 0$
$\implies d \left(d - \frac{4 {v}^{2}}{a}\right) = 0$

Roots are $d = 0 \mathmr{and}$

$d - \frac{4 {v}^{2}}{a} = 0$
$d = \frac{4 {v}^{2}}{a}$ .......(3)

Both roots are valid as these are values of $d$ when cars meet. However, catching up after acceleration is given by (3).