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2.Motion in Straight Line
hard
The displacement $x$ of a particle depend on time $t$ as $x = \alpha {t^{^2}} - \beta {t^3}$
Aparticle will return to its starting point after time $\frac{\alpha }{\beta }$
Bthe particle will come to rest after time $\frac{2\alpha }{3\beta }$
Cthe initial velocity of the particle was zero but its initial acceleration was not zero.
DAll of the above
Solution
Given, $x=\alpha t^2-\beta t^3$
Particle will return to its starting point when, $x =\alpha t ^2-\beta t ^3=0$
or,
$t=\frac{\alpha}{\beta}$
Velocity $= v =\frac{ dx }{ dt }=2 t \alpha-3 \beta t ^2 \ldots \ldots (1)$
at $t =0, v =0$ so the initial velocity zero.
Acceleration $= a =\frac{ dv }{ dt }=2 \alpha-6 t \beta \ldots \ldots(2)$
at $t =0, a =2 \alpha$ so initial acceleration does not zero.
The particle will come to rest when, 2 t $\alpha-3 \beta t ^2=0$ from $(1)$
or, $t=\frac{2 \alpha}{3 \beta}$
$At , t =\frac{\alpha}{3 \beta}, a =2 \alpha-6 \beta\left(\frac{\alpha}{3 \beta}\right)=0$
so net force $= ma =0$, thus no net force act on the particle when $t =\frac{\alpha}{3 \beta}$
Particle will return to its starting point when, $x =\alpha t ^2-\beta t ^3=0$
or,
$t=\frac{\alpha}{\beta}$
Velocity $= v =\frac{ dx }{ dt }=2 t \alpha-3 \beta t ^2 \ldots \ldots (1)$
at $t =0, v =0$ so the initial velocity zero.
Acceleration $= a =\frac{ dv }{ dt }=2 \alpha-6 t \beta \ldots \ldots(2)$
at $t =0, a =2 \alpha$ so initial acceleration does not zero.
The particle will come to rest when, 2 t $\alpha-3 \beta t ^2=0$ from $(1)$
or, $t=\frac{2 \alpha}{3 \beta}$
$At , t =\frac{\alpha}{3 \beta}, a =2 \alpha-6 \beta\left(\frac{\alpha}{3 \beta}\right)=0$
so net force $= ma =0$, thus no net force act on the particle when $t =\frac{\alpha}{3 \beta}$
Standard 11
Physics
