#### To determine

**To find:** The velocity of the particle at time *t.*

#### Answer

The velocity of the particle at time *t* is −Awsin(wt+δ)_.

#### Explanation

**Given:**

The equation of motion of a particle is s=Acos(wt+δ).

**Derivative rule:**

*Constant Multiple Rule:*

If *c* is a constant and f(x) is a differentiable function, then

ddx[cf(x)]=cddx[f(x)] (1)

**Formula used:** *Chain Rule*

If *g* is differentiable at *x* and *f* is differentiable at g(x), then the composite function F=f∘g defined by F(x)=f(g(x)) is differentiable at *x* and F′ is given by the product

F′(x)=f′(g(x))⋅g′(x) (2)

**Recall:**

If x(t) is the displacement of a particle and the time *t* is in seconds, then the velocity of the particle is v(t)=dxdt.

**Calculation:**

Obtain the velocity of the particle at time *t.*

v(t)=ddt(s(t))=ddt(Acos(wt+δ))

Let g(t)=wt+δ and f(u)=Acosu where u=g(t).

Apply the chain rule as shown in equation (2),

v(t)=f′(g(t))⋅g′(t) (3)

The derivative of f′(g(t)) is computed as follows,

f′(g(t))=f′(u)=ddu(f(u))=ddu(Acosu)=Addu(cosu)

Simplify further,

f′(g(t))=A(−sinu)=−Asinu

Substitute u=wt+δ in the above equation,

f′(g(t))=−Asin(wt+δ)

Thus, the derivative f′(g(t))=−Asin(wt+δ).

The derivative of g(t) is computed as follows,

g′(t)=ddt(wt+δ)=ddt(wt)+ddt(δ)=w+0=w

Thus, the derivative g′(t)=w.

Substitute −Asin(wt+δ) for f′(g(t)) and w for g′(t) in equation (3),

v(t)=−Asin(wt+δ)(w)=−Awsin(wt+δ)

Therefore, the velocity of particle at time *t* is v(t)=−Awsin(wt+δ)_.

#### To determine

**To find:** The time such that the velocity is 0.

#### Answer

The velocity is 0 when t=nπ−δw, *n* is an integer.

#### Explanation

**Calculation:**

From part (a), the velocity of particle is v(t)=−Awsin(wt+δ).

Since the velocity is zero, v(t)=0.

That is, −Awsin(wt+δ)=0.

If A≠0 and w≠0, then −sin(wt+δ)=0.

sin(wt+δ)=0sin−1(sin(wt+δ))=sin−1(0)wt+δ=sin−1(0)

For n∈ℕ, sin−1(0)=nπ.

wt+δ=nπwt=nπ−δt=nπ−δw

Therefore, the velocity is 0 when t=nπ−δw, *n* is an integer.