A particle travels horizontally between two
parallel walls separated by 18.4 m. It moves
toward the opposing wall at a constant rate
of 5.8 m/s. Also, it has an acceleration in the
direction parallel to the walls of 1.4 m/s2 .
To find the time it takes for the particle to reach the opposing wall, you can use the equation of motion:
\(d = v_0t + \frac{1}{2}at^2\)
where:
\(d\) is the distance traveled (18.4 m),
\(v_0\) is the initial velocity (5.8 m/s),
\(a\) is the acceleration (1.4 m/s^2),
and \(t\) is the time.
We will rearrange the equation to solve for \(t\):
\(0 = \frac{1}{2}at^2 + v_0t - d\)
Now we have a quadratic equation in terms of \(t\). We can solve this equation using the quadratic formula:
\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
where:
\(a = \frac{1}{2}a\),
\(b = v_0\),
and \(c = -d\).
Plugging in the known values:
\(a = \frac{1}{2} \times 1.4 = 0.7\),
\(b = 5.8\),
and \(c = -18.4\).
\(t = \frac{-5.8 \pm \sqrt{(5.8)^2 - 4 \times 0.7 \times -18.4}}{2 \times 0.7}\)
Simplifying the equation further:
\(t = \frac{-5.8 \pm \sqrt{33.64 + 51.52}}{1.4}\)
\(t = \frac{-5.8 + \sqrt{85.16}}{1.4} \) or \(t = \frac{-5.8 - \sqrt{85.16}}{1.4}\)
Finally, calculating the values of \(t\):
\(t = \frac{-5.8 + 9.22}{1.4}\) or \(t = \frac{-5.8 - 9.22}{1.4}\)
\(t = \frac{3.42}{1.4}\) or \(t = \frac{-15.02}{1.4}\)
\(t = 2.44\) s or \(t = -10.73\) s
Since time cannot be negative, we can ignore the negative value. Therefore, it takes approximately 2.44 seconds for the particle to reach the opposing wall.