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 $2 a$ $2a$ . Show that, when the police car catches up to the other car, the distance is $d = \frac{4 v^{2}}{a}$ $d=(4v^2)/a$ ?

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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 $2 a$ $2a$ . Show that, when the police car catches up to the other car, the distance is $d = \frac{4 v^{2}}{a}$ $d=(4v^2)/a$ ?

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Answer 1 · 2018-03-08T13:02:40

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

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

For Police car
$d = 0 \times t + \frac{1}{2} (2 a) t^{2}$ $d=0xxt+1/2(2a)t^2$
$\Rightarrow d = a t^{2}$ $=>d=at^2$
$\Rightarrow t = \sqrt{\frac{d}{a}}$ $=>t=sqrt(d/a)$ .....(1)
For car

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

Insert value of $t$ $t$ from (1)

$d = v \sqrt{\frac{d}{a}} + \frac{1}{2} a \frac{d}{a}$ $d=vsqrt(d/a)+1/2ad/a$
$\Rightarrow \frac{d}{2} = v \sqrt{\frac{d}{a}}$ $=>d/2=vsqrt(d/a)$

Squaring both sides we get

${(\frac{d}{2})}^{2} = {(v \sqrt{\frac{d}{a}})}^{2}$ $(d/2)^2=(vsqrt(d/a))^2$
$\Rightarrow d^{2} = \frac{4 v^{2}}{a} d$ $=>d^2=(4v^2)/ad$
$\Rightarrow d^{2} - \frac{4 v^{2}}{a} d = 0$ $=>d^2-(4v^2)/ad=0$
$\Rightarrow d (d - \frac{4 v^{2}}{a}) = 0$ $=>d(d-(4v^2)/a)=0$

Roots are $d = 0 and$ $d=0and$

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

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

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A car with speed v=20 ms−1v = "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=4v2ad=(4v^2)/a?

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