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

Hint: 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 $ .