February 2nd, 2019, 12:47 AM  #1 
Newbie Joined: Aug 2018 From: Suisun City, CA Posts: 7 Thanks: 0  Police catching speeder https://imgur.com/gYz41L9 I'm having trouble getting this answer right. I understand that the answer will come from the two kinematic position equations being equal to each other, but despite my efforts, I cannot get it right. I keep coming up with 11.6s but it is not correct. 
February 2nd, 2019, 01:45 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,684 Thanks: 659 
Loaded image  incomprehensible to me.

February 2nd, 2019, 02:38 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,314 Thanks: 1230 
I get $t=14.8416~s$ if this is the correct answer post back and I'll show what I did. 
February 2nd, 2019, 03:03 PM  #4 
Senior Member Joined: Dec 2015 From: iPhone Posts: 387 Thanks: 61 
How to get that value ?
Last edited by idontknow; February 2nd, 2019 at 03:12 PM. 
February 2nd, 2019, 05:41 PM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,042 Thanks: 815 Math Focus: Wibbly wobbly timeywimey stuff. 
The idea here is that, at some time, both the police car and the speeder have the same position. Calling the origin where the speeder passes the police and denoting position by x the motion is described by $\displaystyle x = vt$ for the speeder, where v = 145 km/h $\displaystyle x = V (t  2) + \dfrac{1}{2} a (t  2)^2$ for the cop, where V = 95 km/h (note that V is not the same as v!!) and a = 2.5 m/s^2. The first thing to do is look at the units. I'd change the speeds to m/s, else we have to put both a and the 2 in terms of km/h^2 and h respectively. Doing that we can continue. The position of both cars is the same when the police catches up, so putting the two x's equal: $\displaystyle vt = V (t  2) + \dfrac{1}{2} a (t  2)^2$ Solving this for t gives me t = 17.318 s (after discarding the negative solution. Quick quiz: what does the negative solution represent?) I don't know how romsek got his answer so I can't comment on it. Dan Last edited by topsquark; February 2nd, 2019 at 06:08 PM. 
February 2nd, 2019, 06:49 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,298 Thanks: 1971 
I agree with romsek's answer, but the official answer is probably rounded to 14.8 s.

February 2nd, 2019, 08:20 PM  #7 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,042 Thanks: 815 Math Focus: Wibbly wobbly timeywimey stuff.  
February 3rd, 2019, 12:35 AM  #8  
Senior Member Joined: Sep 2015 From: USA Posts: 2,314 Thanks: 1230  Quote:
 
February 6th, 2019, 09:35 AM  #9 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
I didn't see left side of problem. Will have to get back.
Last edited by zylo; February 6th, 2019 at 10:27 AM. 
February 6th, 2019, 01:19 PM  #10 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 
Police catch up when: T(VcVp)+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 Solving gives t=1.143 hrs=68.8 secs. Add 2 sec to start gives total 70.8 secs EDIT: ds/dt=Vp+at s=tVp+(1/2)at^2 Last edited by zylo; February 6th, 2019 at 01:31 PM. 

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