What Are the Coordinates of the Circle's Center?

[KMjuniormint5]

The question is:

A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point at coordinates (3.00 m, 3.00 m) with a velocity of -5.00 m/s and an acceleration of +10.0 m/s2. What are the coordinates of the center of the circular path?

I know centripical acceleration is a = (v^2)/r

I know:
ax = 0 m/s^2
ay = 10 m/s^2

and . . .

vx = -5 m/s
vy = 0 m/s

where do i go from here?

 

[Doc Al]

I know centripical acceleration is a = (v^2)/r

What will that allow you to calculate?

I know:
ax = 0 m/s^2
ay = 10 m/s^2

and . . .

vx = -5 m/s
vy = 0 m/s

I assume that the directions were given?

Make use of that centripetal acceleration formula.

 

[KMjuniormint5]

ax = 0 m/s^2
ay = 10 m/s^2

and . . .

vx = -5 m/s
vy = 0 m/s

I am just assuming that from reading from the question . . .is that a safe assumption?

This is the question that I am asked:
A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point at coordinates (3.00 m, 3.00 m) with a velocity of -5.00 m/s and an acceleration of +10.0 m/s2. What are the coordinates of the center of the circular path?

and what I did was find the accel. in the x and y direction and the velocity in the x and y velocity

 

[Doc Al]

The question says "with a velocity of -5.00 m/s and an acceleration of +10.0 m/s2", but what direction? You assume the velocity is in the -x direction, but I see nothing in the problem statement that tells you that.

Perhaps this statement "A particle moves horizontally in uniform circular motion, over a horizontal xy plane" was supposed to read "A particle moves along the x-axis...". (If it's moving in a horizontal xy plane, then both x-axis and y-axis are horizontal.)

OK, let's assume your directions are correct. Now make use of the centripetal acceleration formula. You have v and a; find r.

 

[KMjuniormint5]

ahhhhh . . .there is another piece i left out . . "-5.00 i(hat) m/s and an acceleration of +10.0 j(hat) m/s2 "
. . .that is where I got my information from but even in that case would I still just do a direct plug in? so. . .

10 = 25/r and r having a value of 2.5m?

 

[Doc Al]

ahhhhh . . .there is another piece i left out . . "-5.00 i(hat) m/s and an acceleration of +10.0 j(hat) m/s2 "

That makes all the difference!

. . .that is where I got my information from but even in that case would I still just do a direct plug in? so. . .

10 = 25/r and r having a value of 2.5m?

Right. Now use that to locate the center of the circle. (You know which way the acceleration points.)

 

[KMjuniormint5]

wow i was just making it way too hard. . .thank you so much Doc!