Given an accelleration vector, the velocity vector and the origination point, find v(t) and r(t) for the object. What is going on?

Question

Where am I going wrong with this? I am being told that I am integrating and evaluating incorrectly, however I cannot locate where or what I am doing incorrectly. Also the big question for me is how come the constants are being dropped from the book answer.

I have the following for the problem: Suppose an object moves so that its acceleration is given by a = ‹-3 cos t,-2 sin t, 0›. At time t = 0 the object is at (3, 0, 0) and its velocity vector is ‹0, 2, 0›. Find v(t) and r(t) for the object.

Acceleration vector: a⇀ = ‹-3 cos t,-2 sin t, 0›

Time: t = 0

Initial position: p = (3, 0, 0)

Velocity vector: v⇀ = ‹0, 2, 0›

Formulas:

a(t) = v’(t)

v(t) = r’(t)

Given a(t) = ‹-3 cos t,-2 sin t, 0›

v(t) = ∫ a(t) dt

v(t) = ‹ ∫(-3 cos t) dt, ∫(-2 sin t) dt, ∫ 0 dt›

v(t) = ‹ (-3 sin t +C1), (-2 cos t + C2), C3›

v(t) = ‹ (-3 sin t), (-2 cos t), 0› +C⇀

Using initial velocity:

V(0) = C⇀: ‹0, 2, 0›

C1=0

C2=2

C3=0

v(0) = ‹ (-3 sin(0) +0), (-2 cos(0)+ 2), 0+0› = ‹0, 2, 0› (check work)

v(t) = ‹-3 sin(t), (-2 cos(t)+ 2), 0›

Given v(t) = ‹-3 sin(t), (-2 cos(t)+ 2), 0›

r(t) = ∫ v(t) dt

r(t) = ‹∫(-3 sin t) dt, ∫(-2 cos t+2) dt, ∫ 0 dt›

r(t) = ‹ (-3 cos t +C1), (-2 sin t +2t + C2), C3›

r(t) = ‹ (-3 sin t), (-2 sin t +2t), 0› +C⇀

Using initial position:

r(0) = C⇀: ‹3, 0, 0›

C1=3

C2=0

C3=0

r(0) = ‹ (-3 cos(0) +3), (-2 sin(0) +2(0) + 0), 0› = ‹3, 0, 0› (check work)

r(t) = ‹ (-3 sin t + 3), (-2 sin t +2t), 0›

The book answer:

v(t) = ‹-3 sin(t), (2 cos(t)), 0›

Why are the c1, c2 and c3 being dropped? What happens to the 2 for the y position?

r(t) = ‹3 cos(t), (2 sin(t)), 0›

What happens to the + 3 for the x position and the 2t for the y position?

Another problem: Suppose an object moves so that its acceleration is given by a = ⟨−3 cos t, −2 sin t, 0⟩. At time t = 0 the object is at (3, 0, 0) and its velocity vector is ⟨0, 2.1, 0⟩. Find v(t) and r(t) for the object.

The book answer is:

v(t) = ⟨−3sin(t), 2cos(t) + 1/10, 0⟩ Why does this problem only keep the .1 and not the 2?

r(t) = ⟨3cost, 2sint + t/10, 0⟩ Where does the + 3 go for the x position?

Answer 1

1. $\int sin \,x\,dx = -cos\,x + C$
2. Then, using the initial velocity, it is not true that $C_2=2$. Instead, in your case: $$2cos(0)+C_2 = 2$$ thus $$C_2 = 0$$
3. Similarly, using the initial position: $$3cos(0)+C_1 = 3$$ thus $$C_1=0$$

Just for example, your second problem - the velocity in $Y$ component: $$v_y(t)=\int a_y(t)=\int -2sin(t)dt=2cos(t)+c$$ For $t=0$: $$2cos(0)+c=2.1$$ $$c=0.1$$