Expert: Alon Mandes - 3/8/2010

Question
QUESTION: A subatomic particle is traveling in a linear accelerator between time 0 nanoseconds and time 10 nanoseconds, i.e., 0 less than or equal to t and t is less than and equal to 10. Its position is given by s(t)=e^(2t)(2t^2-10t+11) where s  is in meters.

a) At what values of t is the particle at rest? Give your answer as a set of values in braces, e.g., {12,45,34,122}
b) When is the particle moving forward? Give your answer in interval notation.
c) When is the particle moving backward? Give your answer in interval notation.
d) Find, and simplify, s''(t)
e) Is the particle speeding up or slowing down in the postive direction when t=2

ANSWER: a) The particle at rest when the velocity is zero, meaning s'(t)=0 :
  s'(t)=2e^(2t)*(2t^2-10t+11)+e^(2t)(4t-10) . s'(t)=0 --> 2t^2-10t+11=2t-5 -->
  2t^2-12t+16=0 --> t^2-6t+8=0 . Solutions : t1=4 & t2=2 . The answer is {2,4}
b) Moving forward means s'(t)>0 :  0<t<2 OR 4<t<10
c) Moving backward means s'(t)<0 : 2<t<4
d) s''(t)=4e^(2t)*(2t^2-10t+11)+2e^(2t)(4t-10)+2e^(2t)(4t-10)+4e^(2t)
        =2e^(2t)*[4t^2-20t+22+8t-20+2]
        =2e^(2t)*[4t^2-12t+24]
        =8e^(2t)*[t^2-3t+6]
e) s''(t) represent accelaration. s''(2)=32e^(8). This value is positive which means that
  the particle is speeding up .

Alon.


---------- FOLLOW-UP ----------

QUESTION: the answers are not correct......also part b and c need to be in interval notation and part e is slowing down....can u help me figure tht out plz

Answer
Yes Lynn , sorry , there is a mistake :
The solution of s'(t)=0 is :
a) 2t^2-10t+11=-2t+5 --> t^2-4t+6=0 . Solutions : t1=1 & t2=3 . The answer is {1,3}
b) t belongs to the interval nonation : (0,1) OR (3,10)
c) t belongs to the interval nonation : (1,3)
d) s'(t)=2e^(2t)*(2t^2-8t+6) . Therefore :
  s''(t)=4e^(2t)*(2t^2-8t+6)+2e^(2t)*(4t-8)
  s''(t)=4e^(2t)*(2t^2-6t+2)
e) At t=2 : s''(2)=4e^(4)*(2*(4)-6*2+2)=-4e^4 . The value is NEGATIVE therefore the particle
  is slowing down .

My apologies !

Alon.