### Variable Acceleration Motion

• Integrate acceleration function to find velocity and position vectors or displacement.

### Solved Example:

#### 28-4-01

The velocity time relationship is described by the equation $v = P + Qt^2$. The body is travelling with:

Solution:
\begin{align*} v &= P + Qt^2\\ a &= \dfrac{dv}{dt} = 2Qt \end{align*} Hence acceleration depends upon time and hence it is not constant.(non-uniform).

### Solved Example:

#### 28-4-02

An object, moving with a speed of 6.25 $ms^{-1}$, is decelerated at a rate given by: $\dfrac{dv}{dt} = -2.5 \sqrt{v}$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be: (JEE Main Physics 2011)

Solution:
$\dfrac{dv}{dt} = - 2.5 \sqrt{v}$ $\dfrac{dv}{\sqrt{v}} = - 2.5 dt$ Integrating both sides, $2 \sqrt{v} = -2.5 t + C$ Applying boundary condition,
At t = 0, v = 6.25 $C = 5$ Substituting, $2 \sqrt{v} = -2.5 t + 5$ when v = 0 , t= ? $0 = -2.5 t + 5$ $t = 2\ sec$

### Solved Example:

#### 28-4-03

A particle of mass 1 kg is acted upon by a force which varies as shown in the figure. If initial velocity of the particle is 10 m/s, determine what is the maximum velocity attained by the particle. Solution:
The first part of motion is under variable acceleration as the force itself is variable. $v_A = v_0 + \mathrm{Area\ of\ triangle} = 10 + \dfrac{1}{2} \times 10 \times 20 = 110\ m/s$ The next part of the motion is under negative force.
Hence, the velocity will go on decreasing.
Hence, maximum velocity will NOT be affected.

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v$_0$ The distance travelled by the particle in time t will be: