The displacement (\(\Delta x\)) is approximately 0.946 meters.
To find the displacement (\(\Delta x\)), we can use the kinematic equation that relates initial velocity (\(v_0\)), final velocity (\(v\)), acceleration (\(a\)), and displacement (\(\Delta x\)):
\[v = v_0 + at\]
Where:
\(v\) = final velocity = 6.42 m/s
\(v_0\) = initial velocity = 2.35 m/s
\(a\) = acceleration = \(\frac{\Delta v}{\Delta t}\) (change in velocity divided by the change in time)
\(\Delta t\) = time interval = 0.215 s
First, find the acceleration (\(a\)) using the given information:
\[\Delta v = v - v_0 = 6.42 \, \text{m/s} - 2.35 \, \text{m/s} = 4.07 \, \text{m/s}\]
\[a = \frac{\Delta v}{\Delta t} = \frac{4.07 \, \text{m/s}}{0.215 \, \text{s}} \approx 18.884 \, \text{m/s}^2\]
Now, we can use another kinematic equation to find the displacement (\(\Delta x\)):
\[v^2 = v_0^2 + 2a \Delta x\]
Substitute the known values:
\[(6.42 \, \text{m/s})^2 = (2.35 \, \text{m/s})^2 + 2(18.884 \, \text{m/s}^2) \Delta x\]
Solve for \(\Delta x\):
\[41.2164 \, \text{m}^2/\text{s}^2 = 5.5225 \, \text{m}^2/\text{s}^2 + 37.768 \, \text{m/s}^2 \Delta x\]
\[37.6939 \, \text{m/s}^2 \Delta x = 35.6939 \, \text{m}^2/\text{s}^2\]
\[\Delta x = \frac{35.6939 \, \text{m}^2/\text{s}^2}{37.6939 \, \text{m/s}^2} \approx 0.946 \, \text{m}\]
Therefore, the displacement (\(\Delta x\)) is approximately 0.946 meters.