Question 121: Motion in 2D & Vectors

Question

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 :

Accepted Answer

Given that,
x = a cos $$\omega $$t
y = a sin $$\omega $$t
z = a $$\omega $$t
Velocity in x-direction,
V
x
= $${{dx} \over {dt}} =  - a\omega \sin \omega t$$
Velocity  in y-direction,
V
y
= $${{dy} \over {dt}}$$ = a $$\omega $$cos $$\omega $$t
Velocity in z-direction,
V
z
= $${{dz} \over {dt}}$$  = a$$\omega $$
Net velocity,
$$\overrightarrow V $$ = V
x
$$\widehat i$$ + V
y
$$\widehat j$$ + V
z
$$\widehat k$$
Speed = $$\left| {\overrightarrow V } \right| = \sqrt {V_x^2 + V_y^2 + V_z^2} $$
$$ = \sqrt {{a^2}{\omega ^2}{{\sin }^2}\omega t + {a^2}{\omega ^2}{{\cos }^2}\omega t + {a^2}{\omega ^2}} $$
$$ = \sqrt {{a^2}{\omega ^2}\left( {{{\sin }^2}\omega t + {{\cos }^2}\omega t} \right) + {a^2}{\omega ^2}} $$
$$ = \sqrt {2{a^2}{\omega ^2}} $$
$$ = \sqrt 2 a\omega $$

Answer

$$\sqrt 2 \,a\omega $$

Answer

$$a\omega $$

Answer

$$\sqrt 3 \,a\omega $$

Answer

2a$$\omega $$

Motion in 2D & Vectors

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 :

Choose the correct answer: