What is the acceleration and position function of the particle?

Question

x(t) is the position of the particle at time t

x'(t) is its velocity

x''(t) is its acceleration

 

A particle moves on the x-axis at velocity v(t) = 3/(t^.5) t not equal to 0.  At time =1 its position is x=11.

Answer 1

x'(t)=v(t)=3/(t^.5), t≠0
x(t)=∫x'(t)=∫3/(t^.5)=6(t^.5)+C

At t=1, x(t)=11.

x(1)=6(1^.5)+C=11, so C=5

The position function is x(t)=6(t^.5)+5.

x''(t)=v'(t)=-3/(2(t^1.5))

The acceleration function is x''(t)=-3/(2(t^1.5)).

Answer 2

x(t) = ∫v(t)dt = ∫3/t^1/2dt = 3t^1/2/(1/2) + c = 6t^1/2 + c

 

x(1) = 11 = 6(1)^1/2 + c = 6 + c  ⇒ c = 5

 

x(t) = 6t^1/2 + 5

 

a(t) = dv/dt = (-1/2)(3t^-3/2) = (-3/2)/t^3/2