# A 0.27 kg particle moves in an xy plane according to x(t) = -19 + 1t - 2t^3 and y(t) = 19 + 4t -...

## Question:

A 0.27 kg particle moves in an xy plane according to {eq}x(t) = -19 + 1t - 2t^{3} {/eq} and {eq}y(t) = 19 + 4t - 4t^{2} {/eq}, with x and y in meters and t in seconds. At t = 1.1 s, what is the angle (within (-180{eq}^{\circ} {/eq}, 180{eq}^{\circ} {/eq}) interval relative to the positive direction of the x axis) of the net force on the particle?

## Force:

The particle's effect due to its inertia that changes the position or tries to change the position of the particle is called the force. If the direction of the force changes, then the effect of force also changes.

Given Data

• The mass of the particle is: {eq}m = 0.27\;{\rm{kg}} {/eq}.
• The x component of the position of the particle is: {eq}x\left( t \right) = - 19 + t - 2{t^3} {/eq}.
• The y component of the position of the particle is: {eq}y\left( t \right) = 19 + 4t - 4{t^2} {/eq}.
• The time for the particle is: {eq}t = 1.1\;{\rm{s}} {/eq}.

The expression to calculate the x component of velocity of particle is:

{eq}{v_x} = \dfrac{{dx\left( t \right)}}{{dt}} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} {v_x} &= \dfrac{{d\left( { - 19 + t - 2{t^3}} \right)}}{{dt}}\\ {v_x} &= 1 - 6{t^2} \end{align*} {/eq}

The expression to calculate the x component of acceleration of particle is:

{eq}{a_x} = \dfrac{{d{v_x}}}{{dt}} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} {a_x} &= \dfrac{{d\left( {1 - 6{t^2}} \right)}}{{dt}}\\ {a_x} &= - 12t\\ {a_x} &= - 12\left( {1.1\;{\rm{s}}} \right)\\ {a_x} &= - 13.2\;{\rm{m}}/{{\rm{s}}^2} \end{align*} {/eq}

The expression to calculate the y component of velocity of particle is:

{eq}{v_y} = \dfrac{{dy\left( t \right)}}{{dt}} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} {v_y} &= \dfrac{{d\left( {19 + 4t - 4{t^2}} \right)}}{{dt}}\\ {v_y} &= 4 - 8t \end{align*} {/eq}

The expression to calculate the y component of acceleration of particle is:

{eq}{a_y} = \dfrac{{d{v_y}}}{{dt}} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} {a_y} &= \dfrac{{d\left( {4 - 8t} \right)}}{{dt}}\\ {a_y} &= - 8t\\ {a_y} &= - 8\left( {1.1\;{\rm{s}}} \right)\\ {a_y} &= - 8.8\;{\rm{m}}/{{\rm{s}}^2} \end{align*} {/eq}

The expression to calculate the angle relative to positive x axis is:

{eq}\tan \theta = \dfrac{{{a_y}}}{{{a_x}}} {/eq}

Substitute all the values in the above expression.

{eq}\begin{align*} \tan \theta &= \dfrac{{ - 8.8\;{\rm{m}}/{{\rm{s}}^2}}}{{ - 13.2\;{\rm{m}}/{{\rm{s}}^2}}}\\ \theta &\approx 33.7^\circ \end{align*} {/eq}

Thus, theangle relative to positive x axis is {eq}33.7^\circ {/eq}.

Force: Definition and Types

from

Chapter 5 / Lesson 5
78K

Force is everywhere and it comes in a variety of sizes, directions, and types. In this video lesson, you'll identify force as well the different types of force that objects may experience.