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  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126  Quote:
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  $95.0 \, \text{km/hr} = \dfrac{475}{18} \text{m/s}$ $145.0 \, \text{km/hr} = \dfrac{725}{18} \text{m/s}$ $v_s \cdot t = v_{p0} \cdot t + \dfrac{1}{2}a_p (t2)^2$ $\dfrac{725}{18} \cdot t = \dfrac{475}{18} \cdot t + \dfrac{1}{2}(2.50)(t2)^2$ $0 = 9t^2  136t + 36$ $t \approx 14.8 \text{s}$ 
February 7th, 2019, 12:18 PM  #14 
Senior Member Joined: Sep 2015 From: USA Posts: 2,533 Thanks: 1390  
February 7th, 2019, 02:18 PM  #15 
Global Moderator Joined: Dec 2006 Posts: 20,942 Thanks: 2210  

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