Posted by **Sean** on Wednesday, October 13, 2010 at 6:20pm.

A duck has a mass of 2.9 kg. As the duck paddles, a force of 0.07 N acts on it in a direction due east. In addition, the current of the water exerts a force of 0.21 N in a direction of 60° south of east. When these forces begin to act, the velocity of the duck is 0.12 m/s in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in 2.7 s while the forces are acting. How do you get the direction?

- physics -
**Henry**, Friday, October 15, 2010 at 5:55pm
f = ma,

0.07 = 2.9a,

a = 0.07/2.9 = 0.024 m/s^2,

d1 = Vo*t + 0.5*at^2,

= 0.12*2.7 + 0.5*0.024*(2.7)^2,

= 0.3481 + 0.0880 = 0.4361 m., E.

0.21 = 2.9a,

a = 0.21 / 2.9 = 0.0724 m/s^2,

d2 = 0.12*2.7 + 0.5*0.0724*(2.7)^2,

= 0.3481 + 0.2639 = 0.6120 m.@ 60

deg S. of E. = 300 CCW.

RESULTANT(R)

X = hor = 0.4361 + 0.6120*cos300,

= 0.4361 + 0.306 = 0.7421 m.

Y = ver = 0 + 0.6120*sin300 = -0.530m

tanA = y/x = -0.5300 / 0.7421 = -0.7142

A = -35.5 deg = 35.5 deg S.of E. = 324.5 deg CCW.

R = X + iY = 0.7421 - i0.5300

R = X / cosA = 0.7421 / cos324.5 = 0.9115 m @ 35.5 deg. S.of W.

DISPLACEMENT(D)

D = R - Vo*t,

= (0.7421 - i0.5300) - 0.12*2.7

= 0.7421 - 0.324 - i5300,

= 0.4181 - i0.5300,

tanB = Y / X = -0.5300/0.4181=-1.268,

B = -51.7 Deg = 308.3 deg CCW.

Magnitude = x / cosB = 0.4181 / cos308.3 = 0.6751 m.

Direction = 51.7 deg S. of E.

- physics -
**alberto**, Friday, January 28, 2011 at 7:09pm
there is absolutely no way its that complicated. and at points it looks like you made up some values. be more clear.

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