Question
a man walks 5 km due North east, then 4 km due 30 degree N of E and finally 60 degree due S of E for 2 km. What is his final displacement with respect to his starting point ?
Answers
Henry
D = 5km[45o] + 4km[30o] + 2km[-60o].
X = 5*cos45 + 4*cos30 + 2*cos(-60)=8 km.
Y=5*sin45 + 4*sin30 + 2*sin(-60)=3.80km.
tanA = Y/X = 3.80/8 = 0.47544
A = 25.4o
D = 8/cos25.4 = 8.86 km.
X = 5*cos45 + 4*cos30 + 2*cos(-60)=8 km.
Y=5*sin45 + 4*sin30 + 2*sin(-60)=3.80km.
tanA = Y/X = 3.80/8 = 0.47544
A = 25.4o
D = 8/cos25.4 = 8.86 km.