Questions: A particle is moving along a straight line with an initial velocity of 5 m / s when it is subjected to a deceleration of (a=(-1.8 v^1 / 2)) m / s(^2), where (v) is in m / s. Part A Determine how far it travels before it stops. Express your answer to three significant figures and include the appropriate units. Part B How much time does this take? Express your answer to three significant figures and include the appropriate units.

Solution Steps

We are given the deceleration as \(a = -1.8v^{1/2}\). Since \(a = dv/dt\), we can write: \(dv/dt = -1.8v^{1/2}\)

Separate variables and integrate: \(\int_{v_0}^{v} v^{-1/2} dv = \int_{0}^{t} -1.8 dt\)

\(2v^{1/2}\Big|_{v_0}^{v} = -1.8t\Big|_{0}^{t}\)

\(2(v^{1/2} - v_0^{1/2}) = -1.8t\)

Given \(v_0 = 5\) m/s:

\(2(v^{1/2} - \sqrt{5}) = -1.8t\)

\(v^{1/2} = \sqrt{5} - 0.9t\)

\(v = (\sqrt{5} - 0.9t)^2\)

The particle stops when \(v = 0\). \(0 = (\sqrt{5} - 0.9t)^2\) \(\sqrt{5} = 0.9t\) \(t = \frac{\sqrt{5}}{0.9} \approx 2.4845 \text{ s}\)

Since \(v = ds/dt\), we can write: \(ds = v dt\)

Substitute the expression for \(v\) from Step 2: \(ds = (\sqrt{5} - 0.9t)^2 dt\)

Integrate both sides from 0 to \(s\) and 0 to \(t\):

\(\int_0^s ds = \int_0^t (\sqrt{5} - 0.9t)^2 dt\)

\(s = \int_0^t (5 - 1.8\sqrt{5}t + 0.81t^2) dt\)

\(s = 5t - 0.9\sqrt{5}t^2 + 0.27t^3\)

Substitute the time when the particle stops (from Step 3) into the equation for distance:

\(s = 5(2.4845) - 0.9\sqrt{5}(2.4845)^2 + 0.27(2.4845)^3\)

\(s \approx 4.14 \text{ m}\)

Final Answer