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

Explanation

R = \(\sqrt{1^2 + 1^2}\) = \(\sqrt{2km}\)

\(\theta\) = tan-1 \(\frac{(1)}{1}\) = 45oC

= \(\sqrt 2\)km N 45oE

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