Question
For no apparent reason, a poodle is running counter-clockwise 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.
Answers
Damon
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
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