Expert: Paul Klarreich - 4/13/2007

Question
a particle moves with acceleration a= t^1/2 - ( 1/ t^1/2 ) assuming that the velocity v= 4/3 and the position s= 4/15 when t=0, find a) the velocity v in terms of t. b) the position s in terms of t.

Answer
Questioner:   nyawas
Category:  Calculus
 
Subject:  calculus
Question:  a particle moves with acceleration a= t^1/2 - ( 1/ t^1/2 ) assuming that the velocity v= 4/3 and the position s= 4/15 when t=0, find a) the velocity v in terms of t. b) the position s in terms of t.
...................................
Hi, Nyawas,

The scheme here is:

1. Integrate that acceleration to get v(t):

a= t^1/2 - ( 1/ t^1/2 )
a= t^1/2 - t^-1/2

v(t) = t^(3/2)/(3/2) - t^(1/2)/(1/2) + v0, a constant.

v(t) = 2t^(3/2)/3 - 2t^(1/2) + v0

2. Find v0 using your initial condition:  v = 4/3 when t = 0.

v(0) = 2(0)^(3/2)/3 - 2(0)^(1/2) + v0

4/3 = v0

3. Write the velocity function:

v(t) = 2t^(3/2)/3 - 2t^(1/2) + 4/3

Now proceed to the position function, s(t):

1. Integrate the velocity:

s(t) = 2t^(5/2)/[3(5/2)] - 2t^(3/2)/(3/2) + 4t/3 + s0, a constant

s(t) = 4t^(5/2)/15 - 4t^(3/2)/3 + 4t/3 + s0

2. Find s0 using your initial condition:  s = 4/15 when t = 0.

4/15 = s0

3. Write the answer:

s(t) = 4t^(5/2)/15 - 4t^(3/2)/3 + 4t/3 + 4/15