Speeder and Cop Meeting Time Calculation

  • Thread starter Thread starter rainymornings
  • Start date Start date
  • Tags Tags
    Cop
AI Thread Summary
A police car, initially traveling at 80 km/h, is passed by a speeder going 145 km/h, with the police car starting to accelerate 3 seconds later at 2.5 m/s². The key equation used to find the time until the police car overtakes the speeder involves setting their distances equal. Several users attempted calculations, resulting in confusion over whether to add the initial 3 seconds to the final time. Ultimately, the correct approach is to consider the time from when the police car begins accelerating, leading to a final answer of approximately 16.99 seconds after accounting for the initial delay. The discussion highlights the importance of careful calculation and understanding of the problem's wording.
rainymornings
Messages
6
Reaction score
0

Homework Statement


Problem: A police car traveling at 80km/h is passed by speeder traveling at 145km/h. 3 seconds after the speeder passes, the police car starts accelerating at 2.5m/s2. How much time passes before the police overtakes the speeder (moving at a constant speed)?

Variables:
initial velocity of police car (vpi) 80km/h=22.222m/s
initial velocity of speeder (vsi)=final velocity of speeder (vsf) = 40.278m/s
acceleration of police car (ap)=2.5m/s2

Homework Equations


x=x0+v0t+1/2*a*t2

The Attempt at a Solution


I thought that if the distance the police car traveled equals the distance the speeder travelled, that would mean that the police overtook the speeder. So I used the distance formula/equation like this:

time passed since the police started accelerating that the police car and speeder meets=t
vsi*3+vsf*t=vpi*3+vpi*t+1/2*a*t2
40.278*3+40.278t = 22.222*3+22.222*t+1/2*2.5*t2
When I did this equation on mathway I got the answers t=13.0062276,1.4385724.
So if I add 3 seconds to the answers I get t=16 and t=4.
I thought the final answer would be 16 seconds because 4 seconds is too short a time.
But it says my answer is incorrect... Could you tell me where I'm thinking wrong? Thank you!
 
Physics news on Phys.org
Why would you add three seconds to the answer at the end? You already took into account the 3 seconds the policeman was not accelerating.
 
AlephNumbers said:
Why would you add three seconds to the answer at the end? You already took into account the 3 seconds the policeman was not accelerating.

I added 3 seconds because the variable t used in the equation was the time the police car began accelerating. Not since when the speeder passes the police car.
And the answer wasn't 13 seconds either, so I thought it was because I forgot to add 3 seconds :(
 
Relative to the speeder, what is the initial velocity of the police car?
 
Wait I mean relative to the initial velocity of the police car, what is the velocity of the speeder? Ignore my last post.
 
AlephNumbers said:
Wait I mean relative to the initial velocity of the police car, what is the velocity of the speeder? Ignore my last post.

I don't understand what you mean...? According to the question, the initial velocity of the police car is 80km/h and the initial velocity of the speeder is 145km/h.
 
Your solution is correct. It must be a calculation error. Try recalculating.
 
AlephNumbers said:
Your solution is correct. It must be a calculation error. Try recalculating.
I'm constantly getting the numbers wrong:( But thank you for telling me the equations I'm using are right! I feel much better lol
 
When I calculated, I got like 10.2 or something for t. I don't know why our answers are different. Are you using the correct quadratic formula?
 
  • #10
AlephNumbers said:
When I calculated, I got like 10.2 or something for t. I don't know why our answers are different. Are you using the correct quadratic formula?
40.278*3+40.278t = 22.222*3+22.222*t+1/2*2.5*t2
when I simplified this equation I got
1.25t2-18.056t-54.168=0
So I put this into the quadratic formula (-b+√b2-4ac)/2a
and the answer was 16.99...

I'm not really good with the calculations and stuff so I mostly rely on mathway (where they do the quadratic formulas for you)
but the values that come out when I put 40.278 and 22.222 & 80000/3600 and 145000/3600 are quite different
so I am getting extremely confused:(

Thank you so much for your help though.
I'll just try again until I figure the calculations out.
 
  • #11
rainymornings said:
40.278*3+40.278t = 22.222*3+22.222*t+1/2*2.5*t2
when I simplified this equation I got
1.25t2-18.056t-54.168=0
So I put this into the quadratic formula (-b+√b2-4ac)/2a
and the answer was 16.99...

I'm not really good with the calculations and stuff so I mostly rely on mathway (where they do the quadratic formulas for you)
but the values that come out when I put 40.278 and 22.222 & 80000/3600 and 145000/3600 are quite different
so I am getting extremely confused:(

Thank you so much for your help though.
I'll just try again until I figure the calculations out.
I get t = -2.54988 and t = 16.9947 . The negative answer doesn't make sense. Throw it out.

From the wording of the problem, it's not clear whether or not to add the three seconds.

They may be asking for the time from the moment that the police car begins acceleration, not from the time the car passes the police car.
 
  • #12
SammyS said:
I get t = -2.54988 and t = 16.9947 . The negative answer doesn't make sense. Throw it out.

From the wording of the problem, it's not clear whether or not to add the three seconds.

They may be asking for the time from the moment that the police car begins acceleration, not from the time the car passes the police car.

Yes, that's right. I was making calculation mistakes while I was solving this problem. The answer is the value + 3 seconds.
Thank you so much for your help!
 
Back
Top