Question

The displacement of a particle in time t is given by $s = 2{t^2} - 3t + 1$. The acceleration is
(A) 1
(B) 3
(C) 4
(D) 5

Answer 1

Hint: In this question, we need to determine the acceleration of the moving particle of which the displacement is defined by $s = 2{t^2} - 3t + 1$. For this, we will use the relation between the displacement, velocity, and acceleration of the moving particle by carrying out the differentiation process.

Complete step by step answer:
Displacement is the shortest distance between the starting and the endpoint of the journey, whereas distance is the total length of the path traveled by the person.
The differentiation of the distance with respect to time gives us the speed of the moving body, while the differentiation of the displacement with respect to time results in the velocity of the body. Mathematically $\Rightarrow v = \dfrac{{ds}}{{dt}}$.
Here, the equation for the displacement of a moving body has been given as $s = 2{t^2} - 3t + 1$. So, differentiating the displacement equation with respect to time will give us the equation for the velocity of the moving body.
\[
\Rightarrow s = 2{t^2} - 3t + 1 \\
\Rightarrow\dfrac{{ds}}{{dt}} = \dfrac{d}{{ds}}\left( {2{t^2} - 3t + 1} \right) \\
\therefore v = 4t - 3 - - - - (i) \\
 \]
Now, the acceleration of the moving body is the rate of change of velocity of the moving body. Mathematically, $\Rightarrow a = \dfrac{{dv}}{{dt}}$.
Here, the equation for the velocity of the moving body is defined as $v = 4t - 3$. So, differentiating the equation (i) with respect to time results in the acceleration of the moving body.
$
\Rightarrow v = 4t - 3 \\
\Rightarrow \dfrac{{dv}}{{dt}} = \dfrac{d}{{dt}}\left( {4t - 3} \right) \\
\therefore a = 4 \\
$
Hence, the acceleration of the particle of which the displacement is defined as $s = 2{t^2} - 3t + 1$ is given as 4.

Hence,option C is correct.

Note: Students must observe here that we have done the differentiation with respect to time always as the rate of change of the quantity is always given with respect to time. Also,
$Displacement\underset{{{\text{integration}}}}{\overset{{differentiation}}{\longleftrightarrow}}velocity\underset{{{\text{integration}}}}{\overset{{differentiation}}{\longleftrightarrow}}acceleration$