Question

The car on this ramp starts from rest. When released, it

accelerates at a constant rate. It has an initial position of 12 cm
from the top of the ramp, and has an average velocity of 1.20 m/s
for a total of 1.80 s. Which is the correct final position of the car?

Answer 1

Step-by-step explanation:

s 0.12 + 1.20(1.80) = 2.28 m from the top.

Answer 2

The car, starting from rest on a ramp, accelerates with a constant rate. After 1.80 seconds, its final position is 2.28 meters from the initial position, given an average velocity of 1.20 m/s.

To find the final position of the car, you can use the kinematic equation that relates initial position
`(\(s_0\)),` final position
`(\(s\)), initial velocity (\(v_0\)), acceleration (\(a\)), and time (\(t\)):`

`\[ s = s_0 + v_0 t + (1)/(2) a t^2 \]`

Given:

`- \(s_0 = 12 \, \text{cm} = 0.12 \, \text{m}\) (initial position),`

-
`\(v_0 = 0 \, \text{m/s}\)` (initial velocity, as the car starts from rest),

-
`\(a\)` is the acceleration,

-
`\(t = 1.80 \, \text{s}\) (time).`

Since the car starts from rest, the initial velocity
`\(v_0\)` is 0.

The average velocity
`(\(v_{\text{avg}}\))`can be calculated using the formula:

`\[ v_{\text{avg}} = \frac{\text{change in position}}{\text{change in time}} \]`

`\[ v_{\text{avg}} = (s - s_0)/(t) \]`

You know
`\(v_{\text{avg}} = 1.20 \, \text{m/s}\) and \(t = 1.80 \, \text{s}\).` Solve for \(s\):

`\[ s = s_0 + v_{\text{avg}} t \]\[ s = 0.12 \, \text{m} + (1.20 \, \text{m/s})(1.80 \, \text{s}) \]\[ s = 0.12 \, \text{m} + 2.16 \, \text{m} \]\[ s = 2.28 \, \text{m} \]`

So, the correct final position of the car is
`\(2.28 \, \text{m}\)`.