Question

The position of a particle moving along the $X$ -axis is expressed as $x = a{t^3} + b{t^2} + ct + d$ .The initial acceleration of the particle is
(A) $6a$
(B) $2b$
(C) $a + b$
(D) $a + c$

Answer 1

Use the formula of instantaneous acceleration of a particle at a time $t$ . The instantaneous acceleration of a particle is given by, $a = \dfrac{{dv}}{{dt}} = \dfrac{{{d^2}x}}{{d{t^2}}}$ .

Complete answer:
We have here, the expression of the position or displacement at a time $t$ is $x = a{t^3} + b{t^2} + ct + d$ .Now to find the acceleration of the particle for a particular time we first have to differentiate this equation with respect to $t$ , as $t$ is here the independent variable and $x$ is dependant on $t$ .
Now, to differentiate this expression of position, we have to use the formula for differentiation and use it twice two find the expression of acceleration at a particular time $t$ .
The formula for differentiation of ${x^n}$ with respect to $x$ is, $\dfrac{d}{{dx}}{x^n} = n{x^{\left( {n - 1} \right)}}$ .
So, let's find the differentiation of $x$ with respect to $t$ here.
Upon doing that, we first get the expression for velocity as,
$v = \dfrac{d}{{dt}}(x) = \dfrac{d}{{dt}}a{t^3} + \dfrac{{dv}}{{dt}}b{t^2} + \dfrac{{dv}}{{dt}}ct + \dfrac{{dv}}{{dt}}d$
$= a\dfrac{d}{{dt}}{t^3} + b\dfrac{{dv}}{{dt}}{t^2} + c\dfrac{{dv}}{{dt}}t$
Since, differentiation of $\dfrac{d}{{dx}}Cf(x) = C\dfrac{d}{{dx}}f(x)$ where $C$ is any constant
Now, using the formula we get,
$= 3a{t^{\left( {3 - 1} \right)}} + 2b{t^{\left( {2 - 1} \right)}} + 1c{t^{\left( {1 - 1} \right)}}$
On simplifying which becomes,
$= 3a{t^2} + 2b{t^1} + 1c{t^0}$
$= 3a{t^2} + 2bt+ c$ [Since, ${t^0} = 1$ ]
Now on differentiating this expression again we get the expression for acceleration of the particle at an instant of time.
So, differentiating $v$ we get the acceleration $\alpha$ as ,
$\alpha = \dfrac{{dv}}{{dt}} = \dfrac{{{d^2}x}}{{d{t^2}}} = \dfrac{d}{{dt}}(3a{t^2} + 2bt + c)$
Simplifying we get,
$= \dfrac{d}{{dt}}(3a{t^2}) + \dfrac{d}{{dt}}(2bt) + \dfrac{d}{{dt}}(c)$
$= 3a\dfrac{d}{{dt}}{t^2} + (2b\dfrac{d}{{dt}}t) + \dfrac{d}{{dt}}(c)$
Therefore it becomes,
$= 3 \cdot 2a{t^{2 - 1}} + 2 \cdot b{t^{1 - 1}} + 0$
Since, differentiation of a constant is zero, $\dfrac{d}{{dx}}C = 0$
Thus, acceleration of the particle at a instant of time $t$ is,
$\alpha = 6at + 2b$ .
Now, here we have been given to find the initial acceleration of the particle, for that condition time $t$ of the particle is zero.
Therefore, putting the value of time in the expression of the derived acceleration we get,
$\alpha = 6a \cdot 0 + 2b$ .
Which becomes,
$\alpha = 2b$ .
Therefore, the initial acceleration of the particle is $2b$ .
Hence option (B) is correct.

Note:
The acceleration of a particle is generally defined as velocity per time which is $\dfrac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}} = \dfrac{{\Delta v}}{{\Delta t}}$ . Whereas the instantaneous acceleration of the particle is when the time difference tends to zero that means $\mathop {\lim }\limits_{\Delta t \to 0} \Delta t$ .