Differentiation question- how do I find the c

Question

a particle travels along a straight line with its acceleration at the time t secs equal to (3t+2)m/s^2. The particle has an initial positive velocity and travels 30 m in the fourth second. Find the velocity of the body when t=5.

 

 

I got to the point of anti differentiating the acceleration to get the velocity and then displacement:

a=3t+2

v=3/2t^2 +2t+c

x=t^3/2+t^2+ct+d

Answer 1

I am going to use what you have so far.

a=3t+2
v=3/2t² +2t+c
x=t³/2+t²+ct+d

 

The fourth second is the 1 second interval between the 3rd second and the 4th second.

 

Hence, the distance traveled in the fourth second = x(4) -x(3) = 30

32 + 16+ 4c + d - ( 27/2 + 9 + 3c + d) =  30

48 + c - 22.5 = 30

 

From this point you can solve for c , and then for v(5).