Question

The displacement of a particle is given by $x = a_0 + \frac{a_1 t}{2} - \frac{a_2 t^2}{3}$, what is its acceletion ?

• A. $\frac{2a_2}{3}$
• B. $- \frac{2a_2}{3}$
• C. $a_2$
• D. zero

Answer 1

Answer: (B)

Given, $x = a_0 + \frac{a_1 t}{2} - \frac{a_2 t^2}{3}$
Differentiating with respect to $t$, we get
$\frac{dx}{dt} =v = \frac{a_1}{2} - \frac{2a_2 t}{3}$
Again differentiating with respect to $t$, we get
$\frac{d^2 x}{dt^2}= a = -\frac{2a_2}{3}$
$\Rightarrow $ acceleration, $a = -\frac{2a_2}{3}$