For no apparent reason, a poodle is running counter-clockwise at a constant spee

Question

For no apparent reason, a poodle is running counter-clockwise at a constant speed of 4.50m/s in a circle with radius 2.5m . Let change of v 1 be the velocity vector at time t1, and let change of v2 be the velocity vector at time t2. Consider changev =v2 -v1 and changet=t2-t1.

Part A) For changet = 0.7s calculate the magnitude (to four significant figures) of the average acceleration.

Part B)For changet = 0.7s calculate the direction (relative to v? 1) of the average acceleration

Part C)For changet = (4*10)^-2s calculate the magnitude (to four significant figures) of the average acceleration a

Part D) For ?t = (4

Explanation / Answer

PArt A

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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