Question

The coordinates of a particle's motion on a curve is
governed by equations x = - 2 sin t, y = 3 cost,
5 sint. The magnitude of velocity of particle at
time tis
Z=

Answer 1

Given:

The coordinates of a particle's motion on a curve is governed by equations x = - 2 sin(t), y = 3 cos(t)

To find:

Magnitude of Velocity of particle at time t

Calculation:

Velocity function can be obtained by single differentiation of Displacement function;

On x axis:

$x = - 2 \sin(t)$

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

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

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

Again , on y axis:

$y = 3 \cos(t)$

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

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

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

So, Velocity at time t ;

$v_{net} = \sqrt{ ({v_{x})}^{2} + {(v_{y})}^{2} }$

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

$= > v_{net} = \sqrt{ 4 { \cos}^{2} (t) + 9 { \sin}^{2}(t) }$

$= > v_{net} = \sqrt{ 4 \{1 - { \sin}^{2} (t) \} + 9 { \sin}^{2}(t) }$

$= > v_{net} = \sqrt{ 4 + 5 { \sin}^{2} (t) }$

So , final answer is:

$\boxed{ \sf{v_{net} = \sqrt{ 4 + 5 { \sin}^{2} (t) } }}$

Answer 2

Question:-

The coordinates of a particle's motion on a curve is governed by equations x = - 2 sin(t), y = 3 cos(t). The magnitude of velocity of particle at

time t is ?

Answer:-

Velocity in x axis:-

${v}_{x} = \frac{dx}{dt}$

$= > {v}_{x} = \frac{d}{dt} ( - 2 \sin(t) )$

$= > {v}_{x} = - 2 \times \frac{d}{dt} ( \sin(t) )$

$= > {v}_{x} = - 2cos(t)$

Velocity in x axis:-

${v}_{y} = \frac{dy}{dt}$

$= > {v}_{y} = \frac{d}{dt} (3cos(t))$

$= > {v}_{y} = 3 \times \frac{d}{dt} (cos(t))$

$= > {v}_{y} = - 3 \times \sin(t)$

Net Velocity:-

${v}_{net} = \sqrt{ {{v}_{x} }^{2} + {{v}_{y} }^{2} }$

$= > {v}_{net} = \sqrt{ { (- 2cos(t))}^{2} + {( - 3sin(t))}^{2}}$

$= > {v}_{net} = \sqrt{4 {cos}^{2}t + 9 {sin}^{2} t } = \sqrt{4 {cos}^{2}t + 4 {sin}^{2} t + 5 {sin}^{2} t }$

$= > {v}_{net} = \sqrt{4 + 5 {sin}^{2} t}$

$\large{\boxed{\sf{{V}_{net}\:=\:\sqrt{4\:+\:5{sin}^{2}t}}}}$