Expert: Alon Mandes - 11/13/2008

Question
An object moves in space with acceleration a(t)= t(i)+t^2 (j)+cos(2t)k for t >= 0.  If the initial velocity and initial position are v(0)= i+k and r(0)=j respectively. find the position function r(t) of the object.

***okay first thing the "a" and "r" at the beginning of each function both have an arrow over them for vectors.  Second the i,j,k all have the hats on top of them.  and lastly >= means greater than or equal to.  Sorry about the notation but i cant seem to get them on here!

-any help with this would be greatly appreciated thank you so much!

Answer
Ok Sam, We know that the Velocity vector is the integration of
the acceleration vector, so :
V(t)=∫a(t)dt=∫t[i]+t^2[j]+cos(2t)[k]=∫tdt[i]+∫t^2dt[j]+∫cos(2t)dt[k]
V(t)=(1/2)t^2+C1 [i] + (1/3)t^3+C2 [j] + (1/2)sin(2t)+C3 [k]
We need to find the constants C1,C2,C3 from the initial condition
V(t=0)=[i]+[k] -> C1[i] + C2[j] + C3[k] = 1[i] + 0[j] + 1[k], so
C1=C3=1, C2=0 . Thus ,
V(t)=(1/2)t^2+1 [i] + (1/3)t^3 [j] + (1/2)sin(2t)+1 [k] .
Now we will integrate V(t) to ger R(t) :
R(t)=∫(1/2)t^2+1 dt [i] + ∫(1/3)t^3dt [j] + ∫(1/2)sin(2t)+1 dt [k]
R(t)=(1/6)t^3+t+C1 [i] + (1/12)t^4+C2 [j] - (1/4)cos(2t)+t+C3 [k]
Let's use the initial condition R(t=0)=[j] ->
C1[i]+C2[j]-(1/4)+C3[k]=0[i]+1[j]+0[k], so C1=0,C2=1,C3=1/4. Hence
R(t)=(1/6)t^3+t [i] + (1/12)t^4+1 [j] -(1/4)cos(2t)+t+1/4 [k]

Alon.