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2.Motion in Straight Line
easy
The displacement of a particle is proportional to the cube of time elapsed. How does the acceleration of the particle depends on time obtained
A$a \propto {t^2}$
B$a \propto 2t$
C$a \propto {t^3}$
D$a \propto t$
Solution
(d) $x \propto {t^3}$
$\therefore $$x = K{t^3}$
$ \Rightarrow v = \frac{{dx}}{{dt}} = 3\;K{t^2}$ and $ \Rightarrow a = \frac{{dv}}{{dt}} = 6\;Kt$
i.e. $a \propto t$
$\therefore $$x = K{t^3}$
$ \Rightarrow v = \frac{{dx}}{{dt}} = 3\;K{t^2}$ and $ \Rightarrow a = \frac{{dv}}{{dt}} = 6\;Kt$
i.e. $a \propto t$
Standard 11
Physics
