Knights' Collision: Solving a 1D Kinematics Problem for Homework

AI Thread Summary
Two knights start from rest, 62 meters apart, and accelerate towards each other with different magnitudes: Sir George at 0.21 m/s² and Sir Alfred at 0.26 m/s². To find the collision point, the equations for their distances traveled can be set up as X1 = (1/2)(0.21)t² and X2 = (1/2)(0.26)t². The total distance covered by both knights must equal 62 meters, leading to the equation X1 + X2 = 62. By dividing the distance equations, the ratio of their distances can be determined using their respective accelerations. The solution involves finding the time when both knights meet, allowing the calculation of the collision point from Sir George's starting position.
Neil6790
Messages
20
Reaction score
0

Homework Statement


In a historical movie, two knights on horseback start from rest 62 m apart and ride directly toward each other to do battle. Sir George's accleration has a magnitude of 0.21 m/s2, while Sir Alfred's has a magnitude of 0.26 m/s2. How far from Sir George's starting point do the knights collide?

Homework Equations





The Attempt at a Solution


62 - x = (1/2)*(0.21 m/s^2)t^2
62 - x = (1/2)*(0.26 m/s^2)t^2

I was trying to find any velocity or any time but i can't seem to conceive anything logic that would work. Any help would be greatly appreciated.
 
Physics news on Phys.org
Try to write equations that give the position of each knight relative to Sir George's starting position at any given time, and then find when they are equal.
 
Neil6790 said:

Homework Statement


In a historical movie, two knights on horseback start from rest 62 m apart and ride directly toward each other to do battle. Sir George's accleration has a magnitude of 0.21 m/s2, while Sir Alfred's has a magnitude of 0.26 m/s2. How far from Sir George's starting point do the knights collide?

Homework Equations



The Attempt at a Solution


62 - x = (1/2)*(0.21 m/s^2)t^2
62 - x = (1/2)*(0.26 m/s^2)t^2

I was trying to find any velocity or any time but i can't seem to conceive anything logic that would work. Any help would be greatly appreciated.

Almost the right idea.
You need to use different X1 and x2

X1 = 1/2a1t2

X2 = 1/2a2t2

You know that they meet by definition when the time is equal and ...their total distance adds to 62.

Divide the first equation by the second.

You get X1/X2 = a1/a2

and you know X1 + X2 = 62.
 
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