Question

In a given cartesian coordinate system a particle is in the position (initial vector position) ( 9.2 , 3.1 ) meters. After 10 seconds, the particle is in the position (final vector position) ( 72.2, 77.2 ) meters. What is the magnitude of the average velocity of the particle during the given time interval, in m/s

Answer 1

Answer:

Thus, the average velocity is 9.73 m/s.

Explanation:

The velocity is given by the rate of change of position.  

initial position, A = (9.2, 3.1) m

final position, B = (72.2, 77.2) m

time, t = 10 s

The velocity is given by

$\overrightarrow{v} =\frac{\overrightarrow{B}-\overrightarrow{A}}{t}\\\overrightarrow{v}=\frac{(72.2-9.2)\widehat{i} - (77.2-3.1)\widehat{j}}{10}\\\overrightarrow{v}=\frac{63 \widehat{i} - 74.1\widehat{j}}{10}\overrightarrow{v} =6.3 \widehat{i} - 7.41 \widehat{j}\\v =\sqrt{6.3^{2}+7.41^{2}}v = 9.73 m/s$