Particle Straight Line (Integration): Ex Prob 2 A particle experiences a rapid deceleration: a = -2v 3 m/s 2. Initially, at t = 0, v(0) = 8 m/s and s(0)

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Presentation transcript:

Particle Straight Line (Integration): Ex Prob 2 A particle experiences a rapid deceleration: a = -2v 3 m/s 2. Initially, at t = 0, v(0) = 8 m/s and s(0) = 10 m. Please determine: Speed v and position s at t = 4 sec. Discussion: Acceleration is a function! Of velocity! So we must match the acceleration function with one of the defining eqns at right. Our accel function and conditions involve a, v, t and s. Which defining eqn(s) best fit these? Answer: You can use either (1) or (3). If you use (3) now, you can integrate to get an equation relating v and s, but you cannot solve for v or s because you do not have a final position (s) or speed (v). We’ll use (1) now. After integration we will have a v vs. t function. At t = 4 sec (given), we can find v.

A particle experiences a rapid deceleration: a = -2v 3 m/s 2. Initially, at t = 0, v(0) = 8 m/s and s(0) = 10 m. Please determine: Speed v and position s at t = 4 sec. We’ll use (1) now. After integration we will have a v vs. t function…. Defining equation: Separate variables: Set up integrals and sub in limits from conditions.

Set up integrals and sub in limits from conditions. Integrate…. Re-arrange to get v 2 …..

For the second part of this problem, how would you find the position, s, of the particle at t = 4 sec ? You now have two equations to use as starting points….plus the defining eqns. The initial a(v) equation: a = -2v 3 m/s 2. So, which equation can be combined with a defining equation to get position? or the v(t) equation: Answer: You can use either one! The a(v) eqn with (3) (easier) or the v(t) eqn with (2)…(harder). We’ll do both

Hard way: Use the v(t) equation plus defining equation (2)…. We found this v(t) equation…. Solve for v and set equal to ds/dt….. Separate variables and set up the integral…. The integration is difficult. You must use a table or MathCad or other solver to solve it. Fortunately, there is an easier way…. See the next page….

Easier approach: Use the a(v) eqn plus defining equation (3)…. a ds = v dvDefining equation: ( -2v 3 ) ds = v dvSub in our function: Separate variables: Set up integrals and sub in limits from conditions: Integrate: Re-arrange:

Easier approach: Use the a(v) eqn plus defining equation (3)…. At v =.25 m/s…