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 February 6th, 2019, 12:57 PM #11 Global Moderator   Joined: Dec 2006 Posts: 20,942 Thanks: 2210 a=25 m/sec^2=90 km/hr^2 is incorrect. February 7th, 2019, 07:04 AM   #12
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Quote:
 Originally Posted by zylo 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. February 7th, 2019, 10:47 AM #13 Math Team   Joined: Jul 2011 From: Texas Posts: 3,005 Thanks: 1588 $95.0 \, \text{km/hr} = \dfrac{475}{18} \text{m/s}$ $145.0 \, \text{km/hr} = \dfrac{725}{18} \text{m/s}$ $v_s \cdot t = v_{p0} \cdot t + \dfrac{1}{2}a_p (t-2)^2$ $\dfrac{725}{18} \cdot t = \dfrac{475}{18} \cdot t + \dfrac{1}{2}(2.50)(t-2)^2$ $0 = 9t^2 - 136t + 36$ $t \approx 14.8 \text{s}$ February 7th, 2019, 12:18 PM   #14
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Quote:
 Originally Posted by skeeter $t \approx 14.8 \text{s}$
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. February 7th, 2019, 02:18 PM   #15
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Quote:
 Originally Posted by zylo 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. Tags catching, police, speeder Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Colbyelmore Algebra 1 February 21st, 2016 06:44 PM sf7 Algebra 4 April 12th, 2014 07:13 AM

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