Question

A particle starts at the origin and moves away from it in a straight line. After 1.6 seconds, it is 12.8 meters from the origin. It then reverses direction due to a force field with a like charge. After 3.8 seconds total, the particle is at −9.23 m. What is the average velocity of the particle? 3.8 m/s −5.8 m/s −2.4 m/s 13.8 m/s

Answer 1

Answer:

Average velocity of the particle is$-2.4 m/s$

Explanation:

Average velocity is defined as the total velocity divided by the total time. It is given  by the equation$v_av= \frac{displacement}{(total time taken)}$

In the given problem the particle starts from origin, travels $12.8 m$ and then reverses direction and travels till $-9.23 m$ within a total time of $3.8 seconds.$

Displacement is defines as the shortest distance between initial and final point.  

Here displacement = $- 9.23 m$

total time=$3.8 s$

$v_av=-9.23/3.8=-2.4 m/s$