Asked by Candice

A bird watcher meanders through the woods, walking 1.10 km due east, 0.624 km due south, and 1.09 km in a direction 18.2 ° north of west. The time required for this trip is 1.046 h. Determine the magnitudes of the bird watcher's (a) displacement and (b) average velocity.
Answered by Henry
a. D=1.10km - 0.624i km + 1.09km[161.8o]

X = 1.10 + 1.09*cos161.8 = 0.0645 km. =
64.5 m.
Y = - 0.645 + 1.09*sin161.8 = -0.305 km.
= -305 m.

Tan A = Y/X = -305/64.5 = -4.72868
A = -78.06o CW = 78.06o S. of E.

D = Y/Sin(-78.06) = -305/Sin(-78.06) =
312 m.[-78.06].

b. V = D/t = 312m[-78.06]/1.046h = 298
m/h = 0.298km/h[-78.06o]
Answered by Henry
Correction:
a. 1.10km[0o] + 0.624km[270o] + 1.09[161.8o]CCW.

X = 1.10 + 0.624*Cos270 + 1.09*Cos161.8 = 771.10 + 0 - 1.035 = 0.0645km = 64.5 m.

Y = 0.624*sin270 + 1.09*sin161.8 =
-0.624 + 0.340 = -0.284km = -284 m.

Tan A = Y/X = 284/64.5 = -4.40310.
A = -77.2o = 77.2o S. of E. = 282.8oCCW

Displacement = Y/sinA = -284/sin282.8 =
291.2 m[282.8o]CCW.

b. V=(d1+d2+d3)/T=(1.10+.624+1.09)/1.046
= 2.69 km/h.
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