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 $d=(4v^2)/a$ ?

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Answer 1 · 2018-03-08T13:02:40

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

$s=ut+1/2at^2$

Insert value of $t$ from (1)

$d=vsqrt(d/a)+1/2ad/a$
$=>d/2=vsqrt(d/a)$

Squaring both sides we get

$(d/2)^2=(vsqrt(d/a))^2$
$=>d^2=(4v^2)/ad$
$=>d^2-(4v^2)/ad=0$
$=>d(d-(4v^2)/a)=0$

Roots are $d=0and$

$d-(4v^2)/a=0$
$d=(4v^2)/a$ .......(3)

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

A car with speed v = "20 ms"^(-1)v = "20 ms"^(-1) passes a stopped police car. Right at this point, the car accelerates at "2 ms"^-2"2 ms"^-2 and the police car at 2a2a. Show that, when the police car catches up to the other car, the distance is d=(4v^2)/ad=(4v^2)/a?