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2.Motion in Straight Line
medium
The position $x$ of a particle varies with time $t$ as $x = a{t^2} - b{t^3}$. The acceleration of the particle will be zero at time $t$ equal to
A$\frac{a}{b}$
B$\frac{2a}{3b}$
C$\frac{a}{{3b}}$
DZero
(AIPMT-1997)
Solution
$\frac{{dx}}{{dt}} = 2at – 3b{t^2}$
$\Rightarrow \frac{{{d^2}x}}{{d{t^2}}} = 2a – 6bt = 0$
$\Rightarrow t = \frac{a}{{3b}}$
$\Rightarrow \frac{{{d^2}x}}{{d{t^2}}} = 2a – 6bt = 0$
$\Rightarrow t = \frac{a}{{3b}}$
Standard 11
Physics
