A particle moves on a straight line and has acceleration a(t)=18t+14.
Its position at time t=0 is s(0)=6 and its velocity at time t=0 is v(0)=14.
What is its position at time t=7?
a(t)=18t+14
recall that the derivative of your position function yields the velocity
and the derivative of the velocity function will give you the acceleration.
We must integrate:
so v(t) = 9t^2 + 14t + c
given: when t = 0 , v = 14
14 = 0 + 0 + c
c = 14
v(t) = 9t^2 + 14t + 14
s(t) = 3t^3 + 7t^2 + 14t + k
given: when t = 0, s(0) = 6
6 = 0 +0 + 0 + k
k = 6
then s(t) = 3t^3 + 7t^2 + 14t + 6
when t = 7 .....
s(7) = 3(343) + 7(49) + 14(7) + 6
= 1476 , (you gave no units)