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 :

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

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

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

= 4 m

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

For speed in x direction :

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

    =$\frac{d(4cos20t)}{dt}$

    = $-20 \times 4sin20t$

   = -80 sin(20t)

For speed in y direction :

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

    =$\frac{d(4sin20t)}{dt}$

   =$20 \times 4 cos20t$

   = 80 cos(20t)

So,the speed is :

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

=$\sqrt{(-80sin20t)^2+ (80cos20t^2)}$

=$80\sqrt{sin^220t+ cos^220t}$

= 80

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 :

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

    =$\frac{d(-80 sin20t)}{dt}$

    =$-80 \times 20 cos20t$

    =-1600 cos20t

For acceleration in y direction :

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

    =$\frac{d(80cos20t)}{dt}$

   =$-20 \times 80 sin 20 t$

    = -160 sin20t

Direction of acceleration :

$tan\theta = \frac{a_y}{a_x}$

       =$\frac{-1600sin20t}{-1600 cos 20 t}$

       =$\frac{sin20t}{cos20t}$

        = 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 :

$x^2 + y^2 = 16 cos^220t + 16 sin^2 20t$

$x^2 + y^2 = 16$

$x^2 + y^2 = 4$

The sketch of trajectory is  shown in the attached fig .