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Standard XII

Physics

Average Velocity in 2D Motion

A particle st...

Question

A particle starts from rest at the origin and moves along X-axis with acceleration $a = 12 - 2 t$ . The time after which the particle arrives at the origin is (in seconds)

12

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18

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Solution

The correct option is C $18$
Acceleration of particle is given as $a = 12 - 2 t$
Velocity $\int_{o}^{v} d v = \int_{o}^{t} a d t$
Or $\int_{o}^{v} d v = \int_{o}^{t} (12 - 2 t) d t$
Or $v - 0 = 12 t - t^{2}$
We get $v = 12 t - t^{2}$
Position of particle $\int_{o}^{x} d x = \int_{o}^{t} v d t$
Or $\int_{o}^{x} d x = \int_{o}^{t} (12 t - t^{2}) d t$
Or $x = 6 t^{2} - \frac{t^{3}}{3}$
For particle to be at origin, $x = 0$
Or $0 = 6 t^{2} - \frac{t^{3}}{3}$
$⟹ t = 18 s$

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A particle starts from rest at the origin and moves along X-axis with acceleration a=12−2t. The time after which the particle arrives at the origin is (in seconds)

The correct option is C 18Acceleration of particle is given as a=12−2tVelocity ∫vodv=∫toadtOr ∫vodv=∫to(12−2t)dtOr v−0=12t−t2We get v=12t−t2Position of particle ∫xodx=∫tovdtOr ∫xodx=∫to(12t−t2)dtOr x=6t2−t33For particle to be at origin, x=0Or 0=6t2−t33⟹ t=18 s

A particle starts from rest at the origin and moves along X-axis with acceleration $a = 12 - 2 t$ . The time after which the particle arrives at the origin is (in seconds)