HELP: INTEGRATION

G

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the acceleration of an object is given by
a(t)= t + sin(t). If the velocity at t=0 is -5 and teh position of the object at t=0 is 2, find the position of the object as a function of t.

here is what a did so far, but i dont kno what the next step is:

integrate: (t^2)/2 - cost + c = -5 t=0
(0/2) -cos(0) + c= -5
-1 + c = -5
c = -4

THEN WHAT DO I DO??
 
Acceleration is the 2nd derivative of the position function.

You already have the velocity function by integration, \(\displaystyle \frac{t^{2}}{2}-cost-4\).

Integrate again to find the position function.

\(\displaystyle \int({\frac{t^{2}}{2}-cost-4})dt\)

\(\displaystyle =\frac{t^{3}}{6}-sint-4t+C=2\)

Enter t=0 and solve for C
 
SO: (t^3)/6 - sin(t) - 4t + c = -5
0-0-0+ c = -5
c = -5

am i doing this right?? then what should i do?
 
No. My mistake. Should use 2, not -5. Sorry

\(\displaystyle \frac{t^{3}}{6}-sint-4t+C=2\)

\(\displaystyle \frac{(0)^{3}}{6}-sin(0)-4(0)+C=2\)

\(\displaystyle C=2\)

\(\displaystyle \frac{t^{3}}{6}-sint-4t+2\)


To check, take the derivatives.
 
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