A particle moves from an initial position with coordinates x = +7.0 m,y = +2.0 m,to an intermediate position with coordinates x = +4.0 m,y = +5.0 m in a time interval of 3.0 s.The particle then moves from the intermediate position with coordinates x = +4.0 m,y = +5.0 m,to a final position with coordinates x = +1.0 m,y = +8.0 m in a time interval of 6 s.What is the magnitude of the average velocity for the total displacement?
A)0.94 m/s
B)1.1 m/s
C)1.3 m/s
D)2.1 m/s
E)3.0 m/s

Question

Quizplus · Accepted Answer

The average velocity is the displacement divided by the time taken. The displacement is the final position minus the initial position, and the time taken is the total time taken. So, the average velocity is:$$\overrightarrow v_{avg}=\frac{\overrightarrow r_f-\overrightarrow r_i}{t_f-t_i}=\frac{(1\hat{i}+8\hat{j})-(7\hat{i}+2\hat{j})}{6	ext{ s}-3	ext{ s}}=\frac{-6\hat{i}+6\hat{j}}{3	ext{ s}}=-2\hat{i}+2\hat{j}$$The magnitude of the average velocity is:$$|\overrightarrow v_{avg}|=\sqrt{(-2)^2+(2)^2}=\sqrt{8}=\boxed{2.83	ext{ m/s}}$$

A Particle Moves from an Initial Position with Coordinates X