Calculating Distance and Acceleration in Circular Motion of Two Cyclists

AI Thread Summary
Two cyclists, A and B, are moving counterclockwise on a circular track at a constant speed of 8 ft/sec, with cyclist A accelerating at a rate of aA = SA ft/sec². To calculate the distance between the two cyclists after 1 second, one must first determine their initial separation and then account for the distance each travels, considering A's acceleration. Cyclist B maintains a constant speed, moving 8 ft in that time, while the distance A covers includes both its initial speed and the effect of acceleration. The formulas for uniform accelerated motion are essential for these calculations, and the final answers for the distances and accelerations were successfully obtained.
jjiimmyy101
Messages
74
Reaction score
0
Two cyclists A and B, are traveling counterclockwise around a circular track at a constant speed of 8 ft/sec at the instant shown. If the speed of A is increased at aA = SA ft/sec^2, where SA is in ft, determine the distance measured counterclockwise along the track from B to A between the cyclists when time is = 1 sec. What is the magnitude of the acceleration of each cyclist at the instant?

To find the length of an arc, you use the equation arc=theta*radius, but how do you encorporate the time into this? I don't know how far cyclist A moves.

To find the magnitude of acceleration you can use the sqrt of a(normal)^2 + a(tangential)^2.

aB =1.28 ft/sec^2

because a(tangential)=0 (constant velocity) and a(normal)=64/50

I don't know what aA is equal to.

Any suggestions?

I posted a picture too.
 

Attachments

  • circ.gif
    circ.gif
    2.9 KB · Views: 783
Last edited:
Physics news on Phys.org
Originally posted by jjiimmyy101
To find the length of an arc, you use the equation arc=theta*radius, but how do you encorporate the time into this? I don't know how far cyclist A moves.
First find the distance between A & B at time t=0 (the instant shown). Then find out how far each moves in the next second. If it wasn't for that fact that A is accelerating, they would move the same distance, thus maintaining the same separation. But A gains some distance over B: ΔX = 1/2at2.

Also, A gains some speed: ΔV = at.
 
at time t=0 the distance between them is 104.72ft

B moves 8ft in the next second because it is constant

but i still don't get how far A moves.

and how do I use deltaX and deltaV
 
Last edited:
You need to understand the basic formulas for uniform accelerated motion. One key relation is d = v_0t + \frac{1}{2}at^2, which describes the distance traveled in time t. (v_0 is the initial speed.) Another useful formula gives the speed after time t: v = v_0 + at. You will need both of these to understand how "A" moves.

B is just moving at a constant speed (tangential acceleration = 0). I believe you understand that, but note that the above equations apply if you set a=0. (Please try this!)

Of course, the above only applies to the tangential motion. To find the full acceleration, you must add the centripetal acceleration.

Note: Δ just means "change"; ΔX means change in x.
 
Thanks!

I got the final answers.

Thank-you.
 
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 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
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