Question

An object moves in the xy-plane with coordinates: x=4cos(20t)(m) and y=4sin(20t)(m).
a/ Find the object’s distance from the the origin and the speed of the object at every instant of time. Deduce the characteristics of this motion.
b/ Determine the magnitude and the direction of the acceleration vector at each instant of time.
c/ Sketch the trajectory of this object with the velocity vector and acceleration vector at some instant of time t.

Answer 1

Given :

Expression for the x coordinate of the object moving in xy - plane :

x =  4 cos (20t) m

Expression for the y coordinate of the object moving in xy - plane :

y = 4 sin (20t) m

To Find :

a) The distance of object from origin and its speed at every instant of time .

b) The magnitude and direction of the acceleration vector of the object at each instant of time .

c) Sketch of the trajectory with velocity vector and acceleration vector at some instant of time t .

Solution :

a)  

The distance of object from origin will  be :

=$\sqrt{x^2 + y^2}$

=$\sqrt{16cos^2(20t) + 16 sin ^2(20t)}$

$=\sqrt{16[ cos^2(20t) + sin^2(20t)]}$

= 4 m

To get the speed of object we have differentiate the position w.r.t. time , so:

For speed in x direction :

$V_x = \frac{dx}{dt}$

    $= \frac{d(4cos 20t)}{dt}$

    $=-20 \times 4 sin 20 t$

    = -80 sin(20t)

For speed in y direction :

$V_y = \frac{dy}{dt}$

    $= \frac{d(4sin 20t)}{dt}$

     = $20 \times 4 cos 20 t$

      = 80 cos(20t)

So,the speed is :

=$\sqrt{V_x^2 + V_y^2}$

$=\sqrt{(80 cos 20t)^2+ (-80 sin 20t)^2}$

$=80 \sqrt{cos^2 20t + sin^2 20t}$

= 80 $\frac{m}{s}$

And the characteristic of motion is circular motion.

b)

For acceleration we have to differentiate the speed w.r.t time , so :

For acceleration in x direction :

$a_x= \frac{dV_x}{dt}$

   $=d\frac{d(-80sin20t)}{dt}$

   $= -80 \times 20 cos20t$

  =-1600 cos20t

For acceleration in y direction :

$a_y = \frac{dV_y}{dt}$

    $=\frac{d(80 cos 20t)}{dt}$

    $= -20 \times 80 sin20t$

    = -160 sin20t

Direction of acceleration :

$tan\theta =\frac{(\frac{dV_y}{dt})}{(\frac{dV_x}{dt})}$

       $=\frac{-1600 sin 20 t}{-1600 cos 20 t}$

       = tan20t

So, $\theta = 20t$ rad , where $\theta$ is the angle made by the acceleration vector with x axis .

c).

The trajectory of motion of the object is circular with a radius 4 and is given as :

$x^2 + y^2 = 16cos^2 20t + 16 sin^ 2 20 t$

            $=16$

So, $x^2 + y^2 = 4^2$

The sketch of trajectory is  shown in the attached fig .

Answer 2

Given:

An object moves in the xy-plane with coordinates:  x = 4cos(20 t) ( m)  and  y = 4sin(20 t) (m ) .  

To find:

a/ Find the object’s distance from the the origin and the speed of the object at every instant of time. Deduce  the characteristics of this motion.

b/ Determine the magnitude and the direction of the acceleration vector at each instant of time.

c/ Sketch the trajectory of this object with the velocity vector and acceleration vector at some instant of time t

Solution:

From given, we have,

x = 4cos(20 t) ( m)  and  y = 4sin(20 t) (m )

Vx = dx/dt = -80 sin (20t) and Vy = dy/dt = 80 cos (20t)

Ax = dVx/dt = -1600 cos (20t) and Ay = dVy/dt = -1600 sin (20t)

a. The distance from origin,

d = √[x² + y²] = √[16 cos² (20t) + 14 sin² (20t)] = √16 = 4 m

The speed of the object,

Speed = √[Vx² + Vy²] = √[6400 sin² (20t) + 6400 cos² (20t)] = √6400 = 80 m/s

The characteristics of its motion,

The object is moving in a circular path having a radius of 4m moving in counter clockwise direction.

b. The magnitude and the direction of the acceleration

A = √[Ax² + Ay²] = √[(-1600)² cos² (20t) + (-1600)² sin² (20t)] = √1600² = 1600 m/s²

The acceleration is directed towards the origin/centre of the circle.

c. The trajectory of this object with the velocity vector and acceleration vector at some instant of time t is attached below.