Question

A car rolls down a ramp in a parking garage. The horizontal position of the car in meters over time is shown below. Graph of vertical position (in meters) on y axis and time (in seconds) on x axis. The initial position is 12 m at t=0 s, and the position decreases linearly to 6 m at t=8 s. Then the position is constant at 6 m until t=16 s, then decreases linearly to 0 m at t=24 s. \[4\] \[8\] \[12\] \[16\] \[20\] \[24\] \[-15\] \[-12\] \[-9\] \[-6\] \[-3\] \[3\] \[6\] \[9\] \[12\] \[15\] Graph of vertical position (in meters) on y axis and time (in seconds) on x axis. The initial position is 12 m at t=0 s, and the position decreases linearly to 6 m at t=8 s. Then the position is constant at 6 m until t=16 s, then decreases linearly to 0 m at t=24 s. What is the displacement of the car between \[0\text{ s}\] and \[24\text{ s}\]? \[\text m\] What is the distance traveled by the car between \[0\text{ s}\] and \[24\text{ s}\]? \[\text m\] Stuck?Use a hint. Report a problem

Answer 1

To find the displacement of the car between 0 s and 24 s, we need to determine the change in position from the initial position to the final position.

Given:

Initial position = 12 m

Final position = 0 m

Displacement = Final position - Initial position

Displacement = 0 m - 12 m

Displacement = -12 m

Therefore, the displacement of the car between 0 s and 24 s is -12 meters.

To find the distance traveled by the car between 0 s and 24 s, we need to consider the total path length covered by the car. This can be calculated by adding up the distances traveled during each segment of the motion.

Segment 1: The position decreases linearly from 12 m to 6 m over 8 s.

Distance = |Final position - Initial position|

Distance = |6 m - 12 m|

Distance = 6 m

Segment 2: The position remains constant at 6 m for 8 s.

Distance = 6 m

Segment 3: The position decreases linearly from 6 m to 0 m over 8 s.

Distance = |Final position - Initial position|

Distance = |0 m - 6 m|

Distance = 6 m

Total distance = Distance in Segment 1 + Distance in Segment 2 + Distance in Segment 3

Total distance = 6 m + 6 m + 6 m

Total distance = 18 m

Therefore, the distance traveled by the car between 0 s and 24 s is 18 meters.

Answer 2

The displacement of the car between
`\(0 \, \text{s}\)` and
`\(24 \, \text{s}\)` is
`\(-12 \, \text{m}\).`

The distance traveled by the car between
`\(0 \, \text{s}\)` and
`\(24 \, \text{s}\)` is
`\(12 \, \text{m}\).`

How did we get the values?

To find the displacement of the car between
`\(0 \, \text{s}\)` and
`\(24 \, \text{s}\)`, we can use the formula:

`\[ \text{Displacement} = \text{Final Position} - \text{Initial Position} \]`

From the information given, the initial position at
`\(0 \, \text{s}\)` is
`\(12 \, \text{m}\)`, and the final position at
`\(24 \, \text{s}\)` is
`\(0 \, \text{m}\)`. Plug these values into the formula:

`\[ \text{Displacement} = 0 \, \text{m} - 12 \, \text{m} \]`

`\[ \text{Displacement} = -12 \, \text{m} \]`

So, the displacement of the car between
`\(0 \, \text{s}\)` and
`\(24 \, \text{s}\)` is
`\(-12 \, \text{m}\).`

Let's find the distance traveled by the car between
`\(0 \, \text{s}\) and \(24 \, \text{s}\)`. The distance traveled is the total path length covered, which can be found by adding the distances covered in each segment.

1. From
`\(0 \, \text{s}\)` to
`\(8 \, \text{s}\)`: The car moves from
`\(12 \, \text{m}\)` to
`\(6 \, \text{m}\)` linearly, covering a distance of
`\(12 \, \text{m} - 6 \, \text{m} = 6 \, \text{m}\).`

2. From
`\(8 \, \text{s}\)` to
`\(16 \, \text{s}\)`: The car stays at
`\(6 \, \text{m}\)`, so no additional distance is covered.

3. From
`\(16 \, \text{s}\)` to
`\(24 \, \text{s}\)`: The car moves from
`\(6 \, \text{m}\)` to
`\(0 \, \text{m}\)` linearly, covering a distance of
`\(6 \, \text{m}\)`.

Add these distances together:

`\[ \text{Distance} = 6 \, \text{m} + 0 \, \text{m} + 6 \, \text{m} \]`

`\[ \text{Distance} = 12 \, \text{m} \]`

So, the distance traveled by the car between
`\(0 \, \text{s}\)` and
`\(24 \, \text{s}\)` is
`\(12 \, \text{m}\).`