Posted by toshi on Monday, November 28, 2011 at 5:10am.
For no apparent reason, a poodle is running counterclockwise at a constant speed of 3.40 {\rm m/s} in a circle with radius 2.2 {\rm m}. Let \vec v_1 be the velocity vector at time t_1, and let \vec v_2 be the velocity vector at time t_2. Consider \Delta \vec v = \vec v_2 \vec v_1 and \Delta t = t_2  t_1. Recall that \vec a_{\rm av} = \Delta \vec v/ \Delta t.

physics  Damon, Monday, November 28, 2011 at 5:30am
Similar problem done for Jimmy, note different numbers for speed and radius:
geometry if a = 0 at t = 0
angle from x axis (call it a) = v t/r
Vx = v sin a
Vy = v cos a
at start a = 0 so
V1x = 0
V1y = v
at time t
V2x = v sin(vt/r)
V2y = v cos(vt/r)
call change d
d Vx = v sin(vt/r)
d Vy = v[cos(vt/r)1]
ax = (v/t) sin(vt/r)
ay = (v/t)[cos(vt/r)1]
so for our numbers
v/t = 5.4/.7 = 7.714
vt/r = 1.643 about 94 degrees, quadrant2
ax = 7.714 sin(1.643) = 7.694
ay= 7.714[cos(1.643)1] = 8.270
A =SQRT(ax^2+ay^2) = 11.30 m/s^2
compare to v^2/r
v^2/r = 5.4^2/2.3 = 12.7 not too far off centripetal acceleration
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