Question
A dolphin wants to swim directly back to its home bay, which is 0.85 km due west. It can swim at a speed of 4.35 m/s relative to the water, but a uniform water current flows with speed 2.82 m/s in the southeast direction.
(a) What direction should the dolphin head???
(b) How long does it take the dolphin to swim the 0.85-km distance home???
Please helppp
(a) What direction should the dolphin head???
(b) How long does it take the dolphin to swim the 0.85-km distance home???
Please helppp
Answers
4.35m/s @ 180 Deg. + 2.82m/s @ 315 Deg.
X = 4.35*cos180+2.82*cos315=-2.356 m/s.
Y = 4.35*sin180+2.82*sin315=-1.994 m/s.
Q3.
tanAr = Y/X = -1.994 / -2.356= 0.84636,
Ar = 40 Deg.
A = 180 + 40 = 220 Deg.
a. Direction = 180 - 40 = 140 Deg.
b. d = V*t,
t = d/V = 850m / 4.35m/s = 195.4 s. =
3.26 min.
X = 4.35*cos180+2.82*cos315=-2.356 m/s.
Y = 4.35*sin180+2.82*sin315=-1.994 m/s.
Q3.
tanAr = Y/X = -1.994 / -2.356= 0.84636,
Ar = 40 Deg.
A = 180 + 40 = 220 Deg.
a. Direction = 180 - 40 = 140 Deg.
b. d = V*t,
t = d/V = 850m / 4.35m/s = 195.4 s. =
3.26 min.
Vd=4.35m/s @ 180 Deg-2.82m/s @ 315 Deg=
Velocity of dolphin.
X = 4.35*cos180 - 2.82cos315=-6.34 m/s.
Y = 4.35*sin180 - 2.82*sin315=1.994 m/s
Q2.
tanAr = Y/X = 1.994 / -6.34 = -0.31451
Ar = -17.46 Deg. = Reference angle.
A = -17.46 + 180 = 162.5 Deg.
Vd=X/cosA = -6.34 / cos162.5=6.65 m/s
a. Direction = 180 + 17.5 = 197.5 Deg.
b. d = V*t,
t = d/V = 850m / 6.34=134 s.=2.23min.
Velocity of dolphin.
X = 4.35*cos180 - 2.82cos315=-6.34 m/s.
Y = 4.35*sin180 - 2.82*sin315=1.994 m/s
Q2.
tanAr = Y/X = 1.994 / -6.34 = -0.31451
Ar = -17.46 Deg. = Reference angle.
A = -17.46 + 180 = 162.5 Deg.
Vd=X/cosA = -6.34 / cos162.5=6.65 m/s
a. Direction = 180 + 17.5 = 197.5 Deg.
b. d = V*t,
t = d/V = 850m / 6.34=134 s.=2.23min.