Two cars approaching an intersection

  • Thread starter Thread starter inner08
  • Start date Start date
  • Tags Tags
    Cars Intersection
AI Thread Summary
Two cars, A and B, are approaching an intersection from 10 km away, with Car A traveling east at 30 km/h and Car B traveling north at 50 km/h. To determine the minimum distance between the two cars, the distances from each car to the intersection can be expressed as D_A = 10 - 30t and D_B = 10 - 50t. Using the Pythagorean theorem, the distance between the cars can be calculated, and calculus or properties of parabolic functions can be applied to find the time at which this distance is minimized. The closest distance is 3.42 km, occurring when Car A is 2.95 km west and Car B is 1.75 km north of the intersection. This approach effectively combines distance equations and optimization techniques to solve the problem.
inner08
Messages
48
Reaction score
0
Hi,

I have this problem that I can't seem to figure out. I'm taking physics in French but I will try my best to translate it.

At an instant, two cars, A and B, are 10km away from an intersection of two perpendicular roads. Car A is moving towards the East and has a speed of 30km/h whereas car B is moving towards the North at 50km/h (both of them are heading towards the intersection). Determine (a) the distance at the moment where they are the closest to one another; (b) where A and B are located when the distance is minimal?

I already have the answers but I just can't seem to figure out how to get to it.

(a) = 3.42km
(b) = B is 1.75km N and A is 2.95km W of the intersection

Any idea on how to figure this out? I was thinking to use the pythagorean theorem to find the velocity [ (30^2 + 50^2)^1/2 = 58km/h ] and then use that to find the time or something like that. I guess I don't know how to approach this type of problem. Any help would be appreciated!
 
Physics news on Phys.org
If you express the distance from Car A to the intersection as:
D_A = 10 - 30*t (where 30*t is the distance traveled by Car A as it is starting from the point of 10km away from the intersection)

And then express the distance from Car B to the intersection in a similar manner:
D_B = 10 - 50*t

Then by pythagoras expression of these two distances gives the distance between Car A and Car B. One may either use calculus to solve for t, then substitute back in for D, or use properties of parabolic functions (in terms of t) to determine the lowest value of D.
 
Thanks a bunch, I didn't think of doing it that way. :)
 
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