Question

The displacement $x$ of a particle at time $t$ along a straight line is given by $x=\alpha -\beta t+\gamma t^2$. Find the acceleration of the particle.

Answer 1

Step 1: Given

The displacement $x$ of a particle at time $t$: $x=\alpha -\beta t+\gamma t^2$

Step 2: Formula Used

Velocity is the change in displacement per unit of time. So, it is given as $v=\frac{dx}{dt}$, where $x$ is displacement of the body and $t$ is time taken.

Acceleration is the change in velocity per unit of time. So, it is given as $a=\frac{dv}{dt}$, where $v$ is displacement of the body and $t$ is time taken.

Step 3: Solution

Substitute $x=\alpha -\beta t+{\gamma }^2$ in $v=\frac{dx}{dt}$.

$\begin{array}{rcl}v& =& \frac{d}{dt}\left(\alpha -\beta t+{\gamma }^2\right)\\ & =& \frac{d\alpha }{dt}-\frac{d\left(\beta t\right)}{dt}+\frac{d\left(\gamma t^2\right)}{dt}\\ & =& -\beta +2\gamma t\end{array}$

Substitute $\begin{array}{rcl}v& =& -\beta +2\gamma t\end{array}$ for in $a=\frac{dv}{dt}$.

$\begin{array}{rcl}a& =& \frac{d}{dt}\left(-\beta +2\gamma t\right)\\ & =& \frac{d\left(-\beta \right)}{dt}+\frac{d\left(2\gamma t\right)}{dt}\\ & =& 2\gamma \end{array}$

Hence, the acceleration of the particle is $2\gamma$.