Fall+Week+09

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Thursday 10/13
Particle Motion A) v(t)=**x'(t)=sin(t) + 2**
 * Answer to 1985 #2 (back)**

B) A(t)=v'(t)=**x''(t)=cos(t**)

C) Particle is moving to the right when x'(t)>0; sin(t) is positive when t is in the domain (0,**Л)** **Think a little more about this: since all values of sin(t) are between -1 and 1, won't values of sin(t) + 2 be between 1 and 3? Since the interval [1,3] contains all positive numbers, v(t) will be positive no matter what value of t is used. Consequently the particle will move to the right for all values of t. Ms. Gentry**
 * Particle is moving to the right when x'(t)>0 with domain (0,Л)**

D) Distance traveled from t=0 to t=Л/2 is constantly positive, so the particle does not change the direction in which it moves. x(0)= -cos(0)+6=5 x(Л/2)= -cos(Л/2) + Л + 6= Л + 6
 * 1+Л=Total distance traveled**

A) v(t)=x'(t)=1 + ln(t) -3= **ln(t) -2**
 * Answer to 1986 #3 (back)**

B) a(t)=v'(t)=**x''(t)=1/t**

C) position of the particle when it is the most left refers to when the slope is the most negative for a value of t. x'(t)=0=ln(t)-2 ln(t)=2, so e^2=t for x'(t)=0 for values of t smaller than e^2, x'(t) is negative and for values of t greater than e^2, x'(t) is positive **Let's simplify this a little! ln(e^2) = 2ln(e) = 2(1) = 2 so x(e^2) = 2e^2 - 3e^2 + 7 or x(e^2) = 7 - e^2 the farthest left the particle travels. Ms. G**
 * x(t)=t*ln(t)-3t + 7 when t=e^2; the farthest left the particle can go is when x(t)= e^2*ln(e^2)-3*e^2+7**

A) v(t)= tln(t)-t so a(t)=**ln(t)**
 * Answer to 1994 #4 (back)**

B) v(t)=tln(t)-t=0 t=tln(t) 1=ln(t) e=t when v(t)=0 For t'se x*(number > 1)- x which gives a positive velocity So the particle is moving to the right for all **t's>e**

C) The slope of a(t)=0 at a minimum velocity a(t)= ln(t)=0 t=1 because e^0=1 v(1)= 1*ln(1)-1 so the **minimum velocity of the particle is -1**


 * Answer to 1995 #2 (back)**

A) For values of t, with the given domain, the particle moving upward means the slope is positive. Particle is moving upward when x'(t)>0. v(t)=x'(t)=t*cos(t) x'(t)=0 when t=0 or when t=Л/2, 3Л/2, etc. x'(t) is negative when t is between Л/2 and 3Л/2 x'(t) is positive when t is between 0 and Л/2 and between 3Л/2 and 5 therefore, the **particle is moving upwards when t is within the domains (0,Л/2) and (3Л/2, 5].**

B) Acceleration of the particle in terms of t a(t)=v'(t)=x(t) x'(t)=t*cos(t) so x(t)=t*-sin(t)+1*cos(t)=**cos(t)-t*sin(t)=a(t)**

c) Position of the particle the first time the velocity is zero is the value of x(t) when we plug in t for a value that makes the velocity equal to zero. x'(t)=t*cos(t) x'(t)=0 when t=0 and when t=Л/2, 3Л/2, etc. but since we've been told t>0, the **lowest positive value of t to make x'(t)=0 is Л/2.**
 * x(t)=t*sin(t) + cos(t) + 2 for t=Л/2**.
 * Л/2(1) + 0 + 2=Л/2 + 2**

**THANKS SO MUCH FOR POSTING THIS! YOU DID A GREAT JOB! MS. G**