(a) In Problem 48 let s(t) be the distance measured down the inclined plane from the highest point. Use ds/dt = v(t) and the solution for each of the three cases in part (b) of Problem 48 to fi­nd the time that it takes the box to slide completely down the inclined plane. A root-finding application of a CAS may be useful here.

(b) In the case in which there is friction (μ ≠ 0) but no air resistance, explain why the box will not slide down the plane starting from rest from the highest point above ground when the inclination angle u satisfi­es tan θ ≤ μ.

(c) The box will slide downward on the plane when tan θ ≤ μ if it is given an initial velocity v(0) = v₀ > 0. Suppose that μ = √3/4 and θ = 23°. Verify that tan θ ≤ μ. How far will the box slide down the plane if v₀ = 1 ft/s?

(d) Using the values μ = √3/4 and θ = 23°, approximate the smallest initial velocity v₀ that can be given to the box so that, starting at the highest point 50 ft above ground, it will slide completely down the inclined plane. Then ­find the corresponding time it takes to slide down the plane.