Question

A particle moves along the x axis. It is initially at the position 0.180 m, moving with velocity 0.060 m/s and acceleration -0.380 m/s^2. Suppose it moves with constant acceleration for 3.80 s.

Required:
a. Find the position of the particle after this time.
b. Find its velocity at the end of this time interval.

Answer 1

a

`x_2 = -2.3356`

b

`v = -1.384 \ m/s`

Step-by-step explanation:

From the question we are told that

The initial position of the particle is
`x_1 = 0.180 \ m`

The initial velocity of the particle is
`u = 0.060 \ m/s`

The acceleration is
`a = -0.380 \ m/s^2`

The time duration is
`t = 3.80 \ s`

Generally from kinematic equation

`v = u + at`

=>
`v = 0.060 + (-0.380 * 3.80)`

=>
`v = -1.384 \ m/s`

Generally from kinematic equation

`v^2 = u^2 + 2as`

Here s is the distance covered by the particle, so

`(-1.384)^2 = (0.060)^2 + 2(-0.380)* s`

=>
`s = -2.5156 \ m`

Generally the final position of the particle is

`x_2 = x_1 + s`

=>
`x_2 = 0.180 + (-2.5156)`

=>
`x_2 = -2.3356`