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HELP PLEASEEEE>??

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/s2, how much time passes before the police car overtakes the speeder (assumed moving at a constant speed)?

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Answer 1 · 2017-03-08T15:37:22

$6.4 s$ $6.4s$ , rounded to one decimal place.

Explanation:

Let the Police car overtake the speeder after a period $t$ $t$ passes once he crosses the police car.

For first $1.0 s$ $1.0s$ both cars travel at their respective speeds. Distance between the two cars at the end of $1.0 s$ $1.0s$
$d_{1} = Relative speed of cars \times 1.0$ $d_1="Relative speed of cars" xx"1.0"$
$d_{1} = [(100.0 - 80.0) \times \frac{1000}{3600}] \times 1.0 = \frac{20000.0}{3600} = \frac{50}{9} m$ $d_1=[(100.0-80.0)xx1000/3600] xx"1.0"=20000.0/3600=50/9m$
This is the distance speeder is ahead of police car just before it accelerates.
Distance moved by speeder after $1.0 s$ $1.0s$ before Police car overtakes it
$d_{2} = 100.0 \times \frac{1000}{3600} \times (t - 1) = \frac{250}{9} (t - 1) m$ $d_2=100.0xx1000/3600xx(t-1)=250/9(t-1)m$
Distance to be covered by Police car to overtake the speeder in $(t - 1) s$ $(t-1) s$ is
$d_{1} + d_{2} = \frac{50}{9} + \frac{250}{9} (t - 1) m$ $d_1+d_2=50/9+250/9 (t-1)m$
Setting up the kinematic expression
$s = u t + \frac{1}{2} a t^{2}$ $s=ut+1/2at^2$
and inserting given values in SI units we get
$\frac{50}{9} + \frac{250}{9} (t - 1) = \frac{200}{9} (t - 1) + \frac{1}{2} \times 2.0 {(t - 1)}^{2}$ $50/9+250/9 (t-1)=200/9(t-1)+1/2xx2.0(t-1)^2$
$\Rightarrow {(t - 1)}^{2} - \frac{50}{9} (t - 1) - \frac{50}{9} = 0$ $=>(t-1)^2-50/9(t-1)-50/9=0$
$\Rightarrow 9 {(t - 1)}^{2} - 50 (t - 1) - 50 = 0$ $=>9(t-1)^2-50(t-1)-50=0$
To solve the quadratic we substitute $t - 1 = x$ $t-1=x$
It becomes
$9 x^{2} - 50 x - 50 = 0$ $9x^2-50x-50=0$

I found roots of this quadratic using inbuilt graphic tool
my comp my comp

we get roots as $x = - 0.9 and 6.4$ $x=-0.9 and 6.4$ , rounded to one decimal place
Time can not be negative, therefore, ignoring the first root we get

$\Rightarrow t - 1 = 6.4$ $=>t-1=6.4$
$\Rightarrow t = 6.4 s$ $=>t=6.4s$ , rounded to one decimal place.

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