A man travelled \(3 km\) due east in 4 mins by bus on a straight road, and then drove a car \(4 km\) due north in 3 mins. Calculate his average velocity.
A. \(0.71 m / s\) B. \(1.00 m / s\) C. \(11.9 m / s\) D. \(16.7 m / s\)
Correct Answer: C
Explanation
Total displacement \(\bar{d}=3 i km +4 j km\) \(|\bar{d}|=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5 km\) total time \(t=t_{1}+t_{2}\) \(=4 min +3 min =7 min\)verage velocity \(=\frac{\left(\begin{array}{l}\text { total disp- } \\ \text { lacement }\end{array}\right)}{\text { total time }}\) $$ =\frac{5 km }{7 min }=\frac{5 \times 1000}{7 \times 60}=11.9 m / s $$
