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February 2nd, 2019, 12:47 AM   #1
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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.
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February 2nd, 2019, 01:45 PM   #2
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Loaded image - incomprehensible to me.
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February 2nd, 2019, 02:38 PM   #3
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I get $t=14.8416~s$

if this is the correct answer post back and I'll show what I did.
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February 2nd, 2019, 03:03 PM   #4
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How to get that value ?

Last edited by idontknow; February 2nd, 2019 at 03:12 PM.
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February 2nd, 2019, 05:41 PM   #5
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Math Focus: Wibbly wobbly timey-wimey 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.
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February 2nd, 2019, 06:49 PM   #6
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I agree with romsek's answer, but the official answer is probably rounded to 14.8 s.
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February 2nd, 2019, 08:20 PM   #7
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Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by skipjack View Post
I agree with romsek's answer, but the official answer is probably rounded to 14.8 s.
Care to share? I've looked at it again and I can't find an error in my work.

-Dan
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February 3rd, 2019, 12:35 AM   #8
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Quote:
Originally Posted by romsek View Post
I get $t=14.8416~s$

if this is the correct answer post back and I'll show what I did.




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February 6th, 2019, 09:35 AM   #9
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I didn't see left side of problem. Will have to get back.

Last edited by zylo; February 6th, 2019 at 10:27 AM.
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February 6th, 2019, 01:19 PM   #10
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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

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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