JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 25)
The position co-ordinates of a particle moving in a 3-D coordinate system is given by
x = a cos$$\omega $$t
y = a sin$$\omega $$t and
z = a$$\omega $$t
The speed of the particle is :
x = a cos$$\omega $$t
y = a sin$$\omega $$t and
z = a$$\omega $$t
The speed of the particle is :
$$\sqrt 2 \,a\omega $$
$$a\omega $$
$$\sqrt 3 \,a\omega $$
2a$$\omega $$
Explanation
Given that,
x = a cos $$\omega $$t
y = a sin $$\omega $$t
z = a $$\omega $$t
Velocity in x-direction,
Vx = $${{dx} \over {dt}} = - a\omega \sin \omega t$$
Velocity in y-direction,
Vy = $${{dy} \over {dt}}$$ = a $$\omega $$cos $$\omega $$t
Velocity in z-direction,
Vz = $${{dz} \over {dt}}$$ = a$$\omega $$
Net velocity,
$$\overrightarrow V $$ = Vx$$\widehat i$$ + Vy$$\widehat j$$ + Vz$$\widehat k$$
Speed = $$\left| {\overrightarrow V } \right| = \sqrt {V_x^2 + V_y^2 + V_z^2} $$
$$ = \sqrt {{a^2}{\omega ^2}{{\sin }^2}\omega t + {a^2}{\omega ^2}{{\cos }^2}\omega t + {a^2}{\omega ^2}} $$
$$ = \sqrt {{a^2}{\omega ^2}\left( {{{\sin }^2}\omega t + {{\cos }^2}\omega t} \right) + {a^2}{\omega ^2}} $$
$$ = \sqrt {2{a^2}{\omega ^2}} $$
$$ = \sqrt 2 a\omega $$
x = a cos $$\omega $$t
y = a sin $$\omega $$t
z = a $$\omega $$t
Velocity in x-direction,
Vx = $${{dx} \over {dt}} = - a\omega \sin \omega t$$
Velocity in y-direction,
Vy = $${{dy} \over {dt}}$$ = a $$\omega $$cos $$\omega $$t
Velocity in z-direction,
Vz = $${{dz} \over {dt}}$$ = a$$\omega $$
Net velocity,
$$\overrightarrow V $$ = Vx$$\widehat i$$ + Vy$$\widehat j$$ + Vz$$\widehat k$$
Speed = $$\left| {\overrightarrow V } \right| = \sqrt {V_x^2 + V_y^2 + V_z^2} $$
$$ = \sqrt {{a^2}{\omega ^2}{{\sin }^2}\omega t + {a^2}{\omega ^2}{{\cos }^2}\omega t + {a^2}{\omega ^2}} $$
$$ = \sqrt {{a^2}{\omega ^2}\left( {{{\sin }^2}\omega t + {{\cos }^2}\omega t} \right) + {a^2}{\omega ^2}} $$
$$ = \sqrt {2{a^2}{\omega ^2}} $$
$$ = \sqrt 2 a\omega $$
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