Question

Given that the acceleration vector is:
a(t)=(-16cos(4t))i + (-16sin(4t))j + (-1t)k
the initial velocity is: v(0)=i+k, and the initial position vector is: r(0)=i+j+k, compute the velocity vector v(t) and position vector r(t)

Answer 1

a (acceleration) is the second derivative of r (position)
a (acceleration) is the first derivative of v (velocity)

Answer 2

that is
r''=a
r'=v

Answer 3

so to find v
you need to integrate a to find v
and integrate that v to find r

Answer 4

okay so would v(t) = (-4sin(4t)+c1, 4cos(4t)+c2, (-t^2)/2 + c3) ?

Answer 5

ok and v(0)=(1,0,1)

Answer 6

so \[-4\sin(4 \cdot 0)+c_1=1 \\ 4\cos(4 \cdot 0)+c_2=0\\ \frac{-0^2}{2}+c_3=1\]

Answer 7

and now we just solve for c1,2, and 3?

Answer 8

yep

Answer 9

alright, thank you :)

Answer 10

so that will give you the velocity vector

Answer 11

you will need to integrate velocity to find position vector

Answer 12

then apply the condition for position
r(0)=(1,1,1)

Answer 13

shout
if you need more help on this