Verified Solution • 2025–26 Session moderate

QJEEM19Phd1ac: JEE Main 2019 Physics - Class 11 JEEM19Phd1ac Solution: The position co-ordinates of a particle moving in a 3-D…

The position co-ordinates of a particle moving in a 3-D coordinate system is given by
x = a cos ω\omega t
y = a sin ω\omega t and
z = a ω\omega t

The speed of the particle is :

Multiple Choice Options

  • A 2 aω\sqrt 2 \,a\omega
  • B aωa\omega
  • C 3 aω\sqrt 3 \,a\omega
  • D 2a ω\omega

Step-by-Step Verified Solution

Given that,

x = a cos ω\omega t

y = a sin ω\omega t

z = a ω\omega t

Velocity in x-direction,

Vx = dxdt=−aωsin⁡ωt{{dx} \over {dt}} = - a\omega \sin \omega t

Velocity in y-direction,

Vy = dydt{{dy} \over {dt}} = a ω\omega cos ω\omega t

Velocity in z-direction,

Vz = dzdt{{dz} \over {dt}} = a ω\omega

Net velocity,

V→\overrightarrow V = Vx i^\widehat i + Vy j^\widehat j + Vz k^\widehat k

Speed = ∣V→∣=Vx2+Vy2+Vz2\left| {\overrightarrow V } \right| = \sqrt {V_x^2 + V_y^2 + V_z^2}

=a2ω2sin⁡2ωt+a2ω2cos⁡2ωt+a2ω2= \sqrt {{a^2}{\omega ^2}{{\sin }^2}\omega t + {a^2}{\omega ^2}{{\cos }^2}\omega t + {a^2}{\omega ^2}}

=a2ω2(sin⁡2ωt+cos⁡2ωt)+a2ω2= \sqrt {{a^2}{\omega ^2}\left( {{{\sin }^2}\omega t + {{\cos }^2}\omega t} \right) + {a^2}{\omega ^2}}

=2a2ω2= \sqrt {2{a^2}{\omega ^2}}

=2aω= \sqrt 2 a\omega

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