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February 6th, 2019, 01:57 PM   #11
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a=25 m/sec^2=90 km/hr^2 is incorrect.
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February 7th, 2019, 08:04 AM   #12
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Quote:
Originally Posted by zylo View Post
Police catch up when:

T(Vc-Vp)+tVc=tVp+ (1/2)at^2

Vc=145 km/hr
Vp=95 km/hr
T=2 secs=1/30 hr
a=25 m/sec^2=90 km/hr^2 {a=2.5 m/sec^2=9 km/hr^2}

Solving gives

t=1.143 hrs=68.8 secs. {t=11.2 hrs =672 secs}

Add 2 sec to start gives total 70.8 secs {674 secs}


EDIT:

ds/dt=Vp+at
s=tVp+(1/2)at^2
Revision in curly brackets for a=2.5m/sec^2.

The OP is virtually impossible to read. Since you say a=25m/sec^2 is wrong, presumably you have access to original and have "reduced it's size" to make it "come out straight."

In any event, a=2.5m/sec^2, my first guess, gave an overtake time of 11 hrs, which didn't make sense.
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February 7th, 2019, 11:47 AM   #13
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$95.0 \, \text{km/hr} = \dfrac{475}{18} \text{m/s}$

$145.0 \, \text{km/hr} = \dfrac{725}{18} \text{m/s}$

$v_s \cdot t = v_{p0} \cdot t + \dfrac{1}{2}a_p (t-2)^2$

$\dfrac{725}{18} \cdot t = \dfrac{475}{18} \cdot t + \dfrac{1}{2}(2.50)(t-2)^2$

$0 = 9t^2 - 136t + 36$

$t \approx 14.8 \text{s}$
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February 7th, 2019, 01:18 PM   #14
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Quote:
Originally Posted by skeeter View Post
$t \approx 14.8 \text{s}$
3 regulars now have come up with the same answer.

future posters please either point out where the derivation of this answer
is incorrect or just stop.

Imo it's friggin rude to just ignore earlier posts in the thread.
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February 7th, 2019, 03:18 PM   #15
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Quote:
Originally Posted by zylo View Post
Revision in curly brackets for a=2.5m/sec^2.
Your revision was a=2.5 m/sec^2=9 km/hr^2, which is also incorrect.
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