Solve for Knights' Collision Point

  • Thread starter Thread starter skud363
  • Start date Start date
  • Tags Tags
    Collision Point
AI Thread Summary
Two knights, Sir George and Sir Alfred, start 81.5 meters apart and ride toward each other with different accelerations of 0.207 m/s² and 0.339 m/s², respectively. To determine their collision point, both knights' displacements must be equal when they meet, despite their differing speeds. The relevant formula to apply is s = ut + 1/2at², where initial velocities are zero since they start from rest. The total distance covered by both knights combined must equal 81.5 meters. Understanding that they will not travel equal distances before colliding is crucial for solving the problem accurately.
skud363
Messages
2
Reaction score
0
If somone could help me out with this problem I would greatly appreciate it.

Thanks in advance.

In an historical movie, two knights on horseback start from rest 81.5 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.207 m/s2, while Sir Alfred's has a magnitude of 0.339 m/s2. Relative to Sir George's starting point, where do the knights collide?
 
Physics news on Phys.org
Hi skud363,
Welcome to PF! We're here to help you out...show what you've done so far, or your ideas on how to start solving.
 
I understand that the formula x=1/2 (vi+vf)t^2 has to be applied. I am able to find the final velocities easily, but the problem is if they are colliding at one certain point, that means that both of their displacements has to be equal. And since I'm not sure of how much time has elapsed since they are both beginning at rest, it's just a matter of rearranging the formula but I'm not sure how to do that. Please reply w/ any suggestions/detailed explanations, etc...
 
Well I have a neat way of thinking of it. I am not 100% sure that it will work, but I am 99.9%. (I am in a grade 11 physics course). Your horses have to run a total of 81.5m altogether to meet with one another. So think of it like this, if you have both horses start at the exact same point, how would you find out how long until their TOTAL distance (horse A distance + horse B distance) = 81.5m

I think you have to work through a few equations to get your answer.

"but the problem is if they are colliding at one certain point, that means that both of their displacements has to be equal" - This is not true, if one is moving faster than the other, it will travel farther than the other before they colide.

Also, use ^ to indicate a power. 0.207m/s^2 etc ;)
 
Last edited:
They are not accelerating at the same rate and therefore they wouldn't have traveled the same distance when they meet.

Here's what you do,
Let Sa be the distance that Alfred has traveled and Sg for George. You know their initial velocities. Now, apply the general formula s = ut + 1/2at^2 for both knights. (How long does each knight take to reach the point where the collide?)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top