v(t)=t^3
s(t) = (4/3)t^4 + c
s(0) =3
3 = 0 + c
s(t) = (4/3)t^3 + 3
s(2) = (4/3)(8)+3
= 41/3
The velocity of a particle at time t is v(t)=t^3. When t=0, the particle is at position x=3. Determine the location of the particle at time t=2.
A)x=4
B)x=7
C)x=8
D)x=16
E)x=19
2 answers
that should have been:
s(t) = (1/4)t^4 + c
...
s(t) = (1/4)t^4 + 3
s(2) = (1/4)(16) + 3
= 7
please ignore the gibberish I had above
s(t) = (1/4)t^4 + c
...
s(t) = (1/4)t^4 + 3
s(2) = (1/4)(16) + 3
= 7
please ignore the gibberish I had above