PLEASE ANSWER AND EXPLAIN PHASE (IV) PART A! The rest of the problem may be nece

Question

PLEASE ANSWER AND EXPLAIN PHASE (IV) PART A! The rest of the problem may be necessary to do to answer but I ONLY NEED THE ANSWER AND EXPLANATION TO PHASE (IV) PART A which says: when velocity is negative does the ODE for dv/dt change? If it does change EXPLAIN WHY & if it doesn't EXPLAIN WHY.

note: the projectile is thrown up from the ground then reaches a max height and then comes back down. also note: the velocity is positive during the phase : dv/dt = -8 - v

Introduction In lecture, we spoke of air resistance on a falling object and we found that the velocity ys. time curve was some form of exponential decay We also set up a coordinate systeni that was somewhat counterintuitive so that the algebraic signs wrould be less cumbersome. Perhaps that was not the best choice. If had to rewrite that introduction to Ch. 2. I think I would have preferred the standard coordinate system: (R) Upward is pasitive, and downward is negative. We still want g to be a positive number, but we will start with Acceleration: "-g (downward) (b) Ground level is y D. (c) Let say we start at ground level, y(0) 0, and we launch an object straight upward with positive irelocity, y' (0) IF THERE WERE NO AIR RESISTANCE, then this would represent ideal (vertical) projectile motion y' (t) =-gt +?? These are the establishel kinetic equations from Calc. 2 & 3. (d) Suppose I choose a slightly less-massive planet than Earth with g-8 m/s? If we launch the projectile from y 0, then we hav: (t)42 + nt The projectile achieves maximum height at (t)08to seconds and since height vs. time is a parabola, the curve is symmetric about this time. Your Task: Tell me what happens if we introduce air resistance! (I) Suppose that air resistance is proportional to speed: du dt In this first phase, velocity weas positive (upward) Our resistance snust be in the opposite direction of velocity, which must ibe downwared. Thus, we don't tneed the absolute valie bars because we know the velocity is positive during this phase dn Let's suppose that we have a fairly thick atmosphere and b 1 dy dt The initial conditioas are: y (0) 0, y (0) +40 m/sec. II) First, see what happens if we have no air resistance. (t)4t2+40 (a) [3 pls. Determine the time when the projectile achieves maximum height AND deter- mine that maximum height DO NOT LEAVE ANY MATHEMATICAL DETAILS OUT! (III) Now suppose that b 1 Use INTEGRATING FACTOR!! (a) 3 pts.] Find v (t) for t 2 0 up untilv0. The projectile will achieve maximum height and then will change direction (downward). (b) (3 pts.) Find the time when the projoctile achieves maximum height AND deternine that maximum height (IV) Ticky part? Now consider what happens when velocity becomes negative (a) 3 pts Does the ODE for dv/dt change? If it does not change, then you must EXPLAIN why If it does, then change it, and again, EXPLAIN why (b) 3 pts Make a plot of v (t) vs. t for the entire trajectory It should start at t -0 atnd end when the projectile returns to the ground. Obviously, you must show me a calculation for when the projectile returns to the ground. Is the graph symmetric with respect to the time when the projectile achieves maximun height? EXPLAIN

Explanation / Answer

from the definition of acceleration

acceleration is the rate of change of velocity

a = dv/dt

if velocity is positive or negative

differentiation is zero(dv/dt)

so whenever velocity is constant acceleration is zero

a = 0

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