A problem of Irodov




If the expression for co-ordinates of position of particle is given, then the problem be best tackled by component method.

Question :

 A point moves in the plane xy such that x = a sin b t  and y = a ( 1- cos b t  ) ,

where  a and  b are +ve constants. Find :

(a) the distance S traversed by the point during the time T ;

(b) the angle between the point's velocity and acceleration vector.


solution :

we shall solve the problem by component method.


the the distance S traversed by the point during the time T  is given by :

S = _{[ ]}^{[ ]}  | V | dt    ( plz note that only speed is enough to calculate distance )

To calculate | V |  we have to find V  ( bold letters mean vector )

To calculate V  , we proceed remembering V = i V_x + j V_y
Where, i and j are unit vectors along X and Y axis.

Now how to calculate  V_x and V_y   ?

It is given,   x = a sin b t =>    V_x = dx / dt =  a  b  cos b t

also,  y = a ( 1- cos b t )  =>   V_y = dy / dt = a  b  sin b t

so,  V = i V_x + j V_y  = i ( a  b  cos b t ) + j ( a  b  sin b t )

so, | V | = ^{[ ]}  [ ( a  b  cos b t ) ² + ( a  b  sin b t ) ² ]  = a  b 

instantaneous speed    V  = a  b 

so,  S =    _{[ 0]}^{[T ]} | V | dt  =  _{[0 ]}^{[ T]}   a  b  dt  = a  b  T

                               S = a  b  T 


(b )

In the same component method, the acceleration of the point is calculated. The x component of acceleration is :

  w_x  = dV_x / dt = d { a  b  cos b t }  / dt   =  -a b ² sin b t

The y component of acceleration is :

w_y  = dV_y / dt = d { a  b  sin b t }  / dt   =  + a b ² cos b t

so that acceleration vector is w = i w_x + j w_y

or, w = i ( -a b ² sin b t ) + j ( a b ² cos b t )


Now , we have velocity vector V  = i ( a  b  cos b t ) + j ( a  b  sin b t )

acceleration vector   w = i ( -a b ² sin b t ) + j ( a b ² cos b t )

Let C be the angle these 2 vectors. Then

w.V = [ i ( -a b ² sin b t ) + j ( a b ² cos b t ) ] . [ i ( a  b  cos b t )
                                                                                     + j ( a b  sin b t ) ]


wV cos C = - a  ²  b ³  sin b t cos b t  + a  ²  b ³  sin b t cos b t = 0

cos C  = 0

= >   C  = 90 degree.