Discussion Forum

 The position vector of a particle  moving in a plane is given by $r= a \cos \omega t \hat{i}+b \sin  \omega t \hat{j}$,  where $\hat{i}$  and $\hat{j}$  are the unit  vectors  along the rectangular axes X and Y ;a,b and $\omega$  are constants  and t is time . The acceleration  of the particle is directed along the vector 

Answer:

Explanation:

Given  , position vector ,

     $r= a \cos \omega t \hat{i}+b \sin  \omega t \hat{j}$ ........(i)

 on differentiating both sides w.r.t t, we get

 velocity  , $\frac{dx}{dt}=v=-a \omega \sin \omega t \hat{i}+b \omega \cos \omega t \hat{j}$

 Again , on differentiating w.r.t t, we get

 acceleration  $\frac{dv}{dt}=a=-a \omega^{2} \cos  \omega t \hat{i}-b \omega^{2} \sin \omega t \hat{j}$

                      $a=-\omega^{2}(a \cos \omega t \hat{i}+b \sin \omega t \hat{j})$

Form Eq. (i) , we get  $a= \omega^{2}(r)= -\omega^{2}r$

 Hence, acceleration is along -r