JAMB - Physics (1978 - No. 1)
A man walks 1km due east and then 1km due north. His displacement is
1 km N15oE
1 km N30oE
\(\sqrt 2\)km N 45oE
\(\sqrt 2\)km N 60oE
2km N 75oE
ব্যাখ্যা
R = \(\sqrt{1^2 + 1^2}\) = \(\sqrt{2km}\)
\(\theta\) = tan-1 \(\frac{(1)}{1}\) = 45oC
= \(\sqrt 2\)km N 45oE
\(\theta\) = tan-1 \(\frac{(1)}{1}\) = 45oC
= \(\sqrt 2\)km N 45oE