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
$\frac{d^{2}x}{d{t}^2}=a=-\frac{2a_2}{3}$
$\Rightarrow$ acceleration, $a=-\frac{2a_2}{3}$