#### To determine

**To show:** The acceleration of the particle a(t)=v(t)dvds and to explain the difference between the derivatives dvdt and dvds.

#### Explanation

**Given:**

Displacement of the particle is s(t), velocity of the particle is v(t) and acceleration of the particle is a(t).

**Proof:**

Note that,

The velocity of the particle is v(t)=dsdt (1)

The acceleration of the particle is a(t)=dvdt (2)

From equation (2),

a(t)=dvdt=dvds⋅dsdt (by chain rule)=dvds⋅v (by (1))=vdvds

Hence, the acceleration of the particle a(t)=vdvds is proved.

The derivative dvdt is the rate of change of the velocity v(t) with respect to the time *t*.

The derivative dvds is the rate of change of the velocity v(t) with respect to the displacement s(t).