Question

The coordinates of a particle's motion on a curve is
-governed by equations x = 2 sint y = 3 cost,
Z = v5 sint. The magnitude of velocity of particle at
time t is
(1) 312 sint unit (2) 3 unit
(3) 3J2 cost unit (4) 3J2 unit
--
2
Jssin​

Answer 1

Given:

The coordinates of a particle's motion on a curve is:

$x = 2 \sin(t)$

$y = 3 \cos(t)$

$z = 5 \sin(t)$

To find:

Velocity of particle at time t;

Calculation:

Velocity along x axis:

$v_{x} = \dfrac{dx}{dt}$

$= > v_{x} = \dfrac{d \{2 \sin(t) \}}{dt}$

$= > v_{x} = 2 \dfrac{d \{ \sin(t) \}}{dt}$

$= > v_{x} = 2 \cos(t)$

Velocity along y axis:

$v_{y} = \dfrac{dy}{dt}$

$= > v_{y} = \dfrac{d \{3 \cos(t) \}}{dt}$

$= > v_{y} = 3\dfrac{d \{ \cos(t) \}}{dt}$

$= > v_{y} = - 3 \sin(t)$

Velocity along z axis:

$v_{z} = \dfrac{dz}{dt}$

$= > v_{z} = \dfrac{d \{5 \sin(t) \}}{dt}$

$= > v_{z} = 5\dfrac{d \{ \sin(t) \}}{dt}$

$= > v_{z} = 5 \cos(t)$

Net Velocity:

$v_{net} = \sqrt{ { \{2 \cos(t) \}}^{2} + { \{ - 3 \sin(t) \}}^{2} + { \{5 \sin(t) \}}^{2} }$

$= > v_{net} = \sqrt{ 29 { \cos}^{2}(t) + 9 { \sin}^{2}(t) }$

$= > v_{net} = \sqrt{ \{9 { \cos}^{2}(t) + 9 { \sin}^{2}(t) \} + 20 { \cos}^{2} (t) }$

$= > v_{net} = \sqrt{ 9 + 20 { \cos}^{2} (t) }$

$= > v_{net} = \sqrt{ 9 + 20 \{1 - { \sin}^{2} (t) \}}$

$= > v_{net} = \sqrt{ 29 - 20{ \sin}^{2} (t) }$

So, final answer is:

$\boxed{ \sf{ v_{net} = \sqrt{ 29 - 20{ \sin}^{2} (t) } }}$