The position of a particle moving along the x-axis is given by x (t) = 4.0 - 3.0t m. (a) At what time does the particle cross the origin? (b) What is the displacement of the particle between t = 3.0 s and t = 6.0 s? (c) What is the average velocity of the particle between 2.0 s and 4.0 s?

a) t = 4/3 s = 1.33 s

b) ∆x = - 9.0 m

c) Velocity v = - 3m/s

Explanation:

Given;

The position function of the particle can be represented by;

x (t) = 4.0 - 3.0t m

a) the time that the particle cross the origin, at this time x (t) = 0 (origin). So;

x (t) = 4.0 - 3.0t = 0

4.0 - 3.0t = 0

3.0t = 4.0

t = 4/3 s = 1.33 s

b) change in displacement between t = 3s and t = 6s

At t = 3s

x (3) = 4 - 3 (3) = - 5 m

At t = 6s

x (6) = 4 - 3 (6) = - 14m

∆x = x (6) - x (3) = - 14 - (-5) = - 14+5 = - 9m

∆x = - 9.0 m

c) velocity is the change in displacement per unit time.

x (2) = 4 - 3 (2) = - 2

x (4) = 4 - 3 (4) = - 8

∆x = - 8 - (-2) = - 6 m

∆t = 4 - 2 = 2 s

v = ∆x/∆t = - 6/2 = - 3 m/s

Velocity v = - 3m/s