Question

The function a(t)=t^(1/2)−t^(−1/2) m/s^2 represents the acceleration of a particle moving along a horizontal axis. At time t=0, its velocity is equal to (4/3)m/s, and its position is equal to (-4/15) m.
Part A: Find the velocity function v(t) of the particle.

Part B: Find the position function s(t) of the particle.

Answer 1

Answer:

see below

Step-by-step explanation:

a(t)=t^(1/2)−t^(−1/2)

We integrate to find the velocity

v(t) = integral t^(1/2)−t^(−1/2) dt

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

          ------------   -    -----------------  + c  where c is the constant of integration

              3/2                   1/2

v(t) = 2/3 t^ 3/2  - 2 t^ 1/2 +c

We find c by letting t=0 since we know the velocity is 4/3 when t=0

v(0) = 2/3 0^ 3/2  - 2 0^ 1/2 +c = 4/3

       0+c =4/3

       c = 4/3

v(t) = 2/3 t^ 3/2  - 2 t^ 1/2 +4/3

To find the position function we need to integrate the velocity

p(t) = integral 2/3 t^ 3/2  - 2 t^ 1/2 +4/3 dt

     2/3 t ^ (3/2 +1)        2 t ^ (1/2 +1)           4/3t

          ------------   -    -----------------  + ------------- + c  

              5/2                   3/2                    1

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

We find c by letting t=0 since we know the position is -4/15 when t=0

p(0) =  4/15 0^ 5/2 - 4/3 0 ^ 3/2 + 4/3*0 +c = -4/15

         0 +c = -4/15

            c = -4/15

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

Answer 2

Answer:

v(t) = ⅔(t^1.5) - 2(t^0.5) + 4/3

s(t) = (4/15)(t^2.5) - (4/3)(t^1.5) + (4/3)t - 4/15

Step-by-step explanation

a(t) = t^½ - t^-½

Integrate for v(t)

(+1 to the power, divide by the new power)

v(t) = ⅔(t^3/2) - 2(t^½)n+ c

At t = 0, v = 4/3

4/3 = 0 + 0 + c

v(t) = ⅔(t^1.5) - 2(t^0.5) + 4/3

Integrate for s(t)

s(t) = (⅔)(⅖)(t^5/2) - 2(2/3)(t^3/2) + (4/3)t + c

s(t) = (4/15)(t^2.5) - (4/3)(t^1.5) + (4/3)t + c

At t = 0, s = -4/15

-4/15 = 0 - 0 + 0 + c

c = -4/15

s(t) = (4/15)(t^2.5) - (4/3)(t^1.5) + (4/3)t - 4/15