A point moves in the plane `y` according to the law `x = A sin omegat, y = B cos omegat`, where`A,B & omega` are positive constant. The equation for trajectroy for path taken by particle is
A point moves in the plane y according to the law x = A sin omegat, y = B cos omegat , where A,B & omega are positive constant. The velocity of particle is given by
A point moves in the plane y according to the law x = A sin omegat, y = B cos omegat , where A,B & omega are positive constant. The acceleration of particle is given by [ :' barr is position vector]
A point moves in the plane xy according to the law x=a sin omega t, y=b cos omegat , where a,b and omega are positive constants. Find : (a) the trajectory equation y(x) of the point and the direction of its motion along this trajectory , (b) the acceleration w of the point as a function of its radius vector r relative to the orgin of coordinates.
A particle moves in xy plane according to the law x = a sin omegat and y = a(1 – cos omega t) where a and omega are constants. The particle traces
A point moves in the plane xy according to the law x=a sin omegat , y=a(1-cos omega t) , where a and omega are positive constants. Find: (a) the distance s traversed by the point during the time tau , (b) the angle between the point's velocity and acceleration vectors.
A point moves in the plane xy according to the law, x=a sin omegat,y=a(1-cosomegat) Answer the following question taking a and omega as positive constant The equation of the trajectory of the particle is
A point moves in the plane xy according to the law x = alpha sin omega t, y = alpha(1 - cos omega t) , where alpha and omega are positive constant and t is time. Find the distance traversed by point in time t_(0) .
A point moves in the plane xy according to the law, x=a sin omegat,y=a(1-cosomegat) Answer the following question taking a and omega as positive constant The magnitude of the velocity of the point as a function of time is
