The motion of a particle along a straight line is described by the function, $x=(2 t-3)^2$ where $x$ is in…
The motion of a particle along a straight line is described by the function, $x=(2 t-3)^2$ where $x$ is in metres and $t$ is in seconds. The acceleration of the particle at $t=2 \mathrm{~s}$ is
$1 \mathrm{~ms}^{-2}$
$4 \mathrm{~ms}^{-2}$
$8 \mathrm{~ms}^{-2}$
$7 \mathrm{~ms}^{-2}$
Solution
Given, $x=(2 t-3)^2$
where,
$
\begin{aligned}
& x=\text { displacement and } t=\text { time. } \\
& x=4 t^2+9-12 t
\end{aligned}
$
Differentiate w.r.t. time, we get
Velocity, $v=\frac{d x}{d t}=8 t-12$
Acceleration, $a=\frac{d v}{d t}=8 \mathrm{~m} / \mathrm{s}^2$
Alternative Solution:
Sure.
The given function is $x = (2t - 3)^2$. This function describes the position of the particle at any given time $t$.
To find the acceleration, we first need to find the velocity, which is the first derivative of the position function, and then the acceleration, which is the derivative of the velocity function.
Let's find the velocity first. The derivative of $x$ with respect to $t$ (denoted as $\frac{dx}{dt}$ or $x'$) is the velocity $v$.
Using the chain rule, the derivative of $(2t - 3)^2$ is $2(2t - 3) \cdot 2 = 4(2t - 3)$.
So, the velocity $v$ is $4(2t - 3)$.
Now, let's find the acceleration, which the derivative of the velocity function. The derivative of $4(2t - 3)$ is simply $8$.
This means the acceleration is $8 \, \mathrm{ms}^{-2}$ and does not depend on $t$.
So, the acceleration of the particle at $t = 2 \, \mathrm{s}$ is also $8 \, \mathrm{ms}^{-2}$.
So, the correct answer is C) $8 \, \mathrm{ms}^{-2}$.