Question

The velocity of a particle is given by, $v=4+2\left( {{C}_{1}}+{{C}_{2}}t \right)$ where ${{C}_{1}},{{C}_{2}}$ are constant. Find initial velocity and acceleration of the particle.

Answer 1

Hint:We are given the velocity of the particle as a function of time. This shows that the particle does not undergo a constant displacement per unit time. The velocity of the particle is different at every instant of time and hence the three equations of motion cannot be applied on it to find its acceleration. Thus, we will differentiate the velocity with respect to time to find acceleration.

Complete answer:
To find initial velocity, put $t=0$ in given equation of instantaneous velocity:
$\begin{align}
  & \Rightarrow v=4+2\left( {{C}_{1}}+{{C}_{2}}.0 \right) \\
 & \Rightarrow v=4+2\left( {{C}_{1}} \right) \\
\end{align}$
Therefore, initial velocity is $4+2{{C}_{1}}$.
To find the acceleration, divide the velocity change over an interval of time with that particular interval of time itself.
Acceleration, $a=\dfrac{\Delta v}{\Delta t}$
$\Rightarrow a=\dfrac{v-u}{t}$
But when a particle has instantaneous velocity, then we calculate instantaneous acceleration by differentiating the instantaneous velocity with respect to time.
$\Rightarrow {{a}_{\text{instantaneous}}}=\dfrac{d{{v}_{\text{instantaneous}}}}{dt}$
We shall now differentiate the equation of velocity with respect to time to find instantaneous acceleration of the particle:
Acceleration, $a=\dfrac{dv}{dt}$
$\begin{align}
  & \Rightarrow a=\dfrac{d}{dt}\left( 4+2\left( {{C}_{1}}+{{C}_{2}}t \right) \right) \\
 & \Rightarrow a=\dfrac{d}{dt}\left( 4+2{{C}_{1}}+2{{C}_{2}}t \right) \\
\end{align}$
Since the derivative of constants is equal to zero and $\dfrac{d}{dx}\left( bx \right)=b$ (where b is constant):
$\Rightarrow a=2{{C}_{2}}$
Therefore, initial velocity of particle having velocity $v=4+2\left( {{C}_{1}}+{{C}_{2}}t \right)$is $4+2{{C}_{1}}$ and the acceleration of the particle is $2{{C}_{2}}$.

Note:
The acceleration of the given particle is not a function of time. Hence, it is constant. This implies that the particle is experiencing uniform acceleration or going under uniformly accelerated motion. The S.I. the unit of acceleration is $m{{s}^{-2}}$. However, the positive acceleration of the particle shows that the velocity of the particle is increasing at the same rate per unit time in the same direction as that of the initial velocity.