Show your work. 1. Starting from rest at time t = 0, a particle begins moving in a circular path of radius 2 In. It is convenient to consider a coordinate system that moves with the object such that the parallel component, denoted , is always in the direction of the instantaneous velocity while the perpendicular component, denoted L, is always at a 90~degree angle to the parallel component. The object's speed as a function of time t is given by v(t) : v\" (t) : a.\" t, for t> 0 with a\" = 0.4 111/82. A. At what time t1 will the object have completed one complete revolution? B. What is the magnitude of the acceleration of the object at time 151? (Hint: Like any vector, you can nd the magnitude by nding its two components rst.) C. At time t1, at what angle is the acceleration vector pointing? Use the convention that the parallel direction is 0 degrees and the perpendicular direction toward the center is +90 degrees. 2. A projectile is launched from the ground at an angle 9 above horizontal. A wall of height H is a horizontal distance D away from the launching location. A. What is the minimum launch speed '0ij that is needed such that the projectile is able to make it over the top of the wall? (In other words, the projectile nearly touches the top of the wall while it goes over.) Your nal answer sin :9 cos 9 as anything else for the rest of this problem. Also, do not assume 113; = 0 when the projectile goes over the wall; will be in terms of H, D, 6, and g. (Suggestion: lfyou get a term, rewrite it as tanH and never rewrite it D. Now let's suppose it is known that the projectile is actually moving downward (By 2H/D. You have thus found a geometric inequality that must be true if the projectile even has a chance of moving downward as it goes over the wall, regardless of the actual launch speed. (Hint: The algebra looks nasty at rst sin 6 cos B but can be simplied greatly. Look for another term and rewrite it as tan (9. This should help simplify.)