Question

A marble is rolled across the floor, given the an initial velocity of 2m/s. The marble slows down due to friction with the floor with an acceleration of -0.5m/s^2. How far does the marble travel before stopping?

Answer 1

4 meters

Step-by-step explanation:

To find the distance traveled by the marble before stopping given that its acceleration is constant, we can use the following kinematic equation:

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

where:

• u is the initial velocity.
• v is the final velocity.
• a is the acceleration.
• s is the displacement (distance traveled).

In this case:

• u = 2 m/s
• v = 0 m/s (since the marble stops)
• a = -0.5 m/s²

Substitute these values into the equation and solve for s:

`\begin{aligned}(0)^2&=(2)^2+2(-0.5)s\\\\0&=4-s\\\\s&=4\end{aligned}`

Therefore, the marble travels 4 meters before stopping.

Answer 2

4 meters

Step-by-step explanation:

To find the distance traveled by the marble before stopping, we can use the kinematic equation that relates initial velocity (
`\sf v_0`), acceleration (
`\sf a`), and displacement (
`\sf d`):

`\sf v^2 = v_0^2 + 2ad`

Where:

-
`\sf v` is the final velocity (which is 0 when the marble stops),

-
`\sf v_0` is the initial velocity,

-
`\sf a` is the acceleration, and

-
`\sf d` is the displacement.

In this case,

• 
  `\sf v = 0`,
• 
  `\sf v_0 = 2 \, \textsf{m/s}`, and
• 
  `\sf a = -0.5 \, \textsf{m/s}^2`

Substitute these values into the equation:

`\sf 0 = (2)^2 + 2(-0.5)d`

Simplify and solve for
`\sf d`:

`\sf 0 = 4 - d`

`\sf d = 4 \, \textsf{meters}`

So, the marble travels
`\sf 4 \, \textsf{meters}` before stopping.