calculus- acceleration question?

Question

The acceleration of an object is given by a(t)=6sin(t) with the initial velocity of -9.5. Find the distance the object travels on the interval [0,pi] to the nearest integer.

Answer 1

distance traveled is integral of absolute value of velocity.
So we must first find v(t) using the knowledge that if you know acceleration its antiderivaitve is velocity.
since a(t) = 6 sint ===> v(t) = - 6 cost + C
now use the given info that at t = 0 v = -9.5
v(t) = - 6 cost + C
- 9.5 = - 6 cos0 + C ===> recall cos0 = 1
-9.5 = -6 + C
C = -3.5
v(t) = - 6 cost - 3.5

Now distance = INT IvI dt from 0 upto pi
INT I- 6 cost - 3.5I dt
using a calculator type in to get answer. Rememebr we can not use formulas for antiderivaitves with absolute values.
**make sure calculator is in RADIAN mode**

TI - key strokes would be
[MATH] [9] to get fnInt
[MATH] [>] to get NUM menu [1] ABS

fnInt ( ABS(-6cost - 3.5) , x , 0 , pi) = 14.107


to the nearest integer the object traveled 14

Answer 2

Question is unclear---is initial velocity 9.5 or - 9.5?
Will assume 9.5 but working would be the same.
a = d²x/dt² = 6 sin t
v = dx/dt = - 6.cos t + C
v = - 6.cos t + 9.5
d = - 6 sint + 9.5t + c
When t = 0 ,d = 0 ,c = 0
d = - 6.sin t + 9.5t-------limits 0 to π
d = 9.5 π
d = 30 (to nearest integer)