Question

.A particle starts on the origin. It is pushed back to -5.7 m in 2.1 s. Then it is pushed

forward to 7.3 m. The entire motion of the particle takes 9.5 seconds. What is the
average velocity of the particle?

Answer 1

The average velocity of the particle is 1.76 m/s.

Step-by-step explanation:

Given that,

Distance = -5.7 m

Time t = 2.1 s

Forward distance = 7.3 m

Total time = 9.5 sec

We need to calculate the average velocity of the particle

Average velocity :

Average velocity is equal to the displacement divided by change in time.

`v = (\Delta D)/(\Delta t)`

Where, D = displacement

t = change in time

`v=(7.3-(-5.7))/(9.5-2.1)`

`v=1.76\ m/s`

Hence, The average velocity of the particle is 1.76 m/s.

Answer 2

Answer: The average velocity is -0.965m/s

Explanation: The first step is to calculate the two velocities is both directions. A velocity is a distance per unit time.

V=d/ t

=-5.7/2.1

=-2.7m/s

For the other direction the velocity is

V=7.3/9.5

=0.77m/s

The average velocity the add the velocities and divide them by 2.

V=-2.7+0.77/2

V= 0.965m/s