A man walks 1km due east and then 1km due north. His displacement is
A
1 km N15oE
B
1 km N30oE
C
\(\sqrt 2\)km N 45oE
D
\(\sqrt 2\)km N 60oE
E
2km N 75oE
Verified Correct Answer: C.
Physical Principles & Numerical Solution
R = \(\sqrt{1^2 + 1^2}\) = \(\sqrt{2km}\)
\(\theta\) = tan-1 \(\frac{(1)}{1}\) = 45oC
= \(\sqrt 2\)km N 45oE
Source: Myschool Past Questions Archive (Record #29231)