A particle of mass `2sqrt(5) Kg` is having velocity `vec v=(2hat i+hat j)m/s`" and acceleration `vec a=(hat j-hat i)m/s^(2)`" at some instant of time. At same instant find component of net force (in N) parallel to velocity.

A particle starts moving with velocity "vecV=hat i(m/s)", and acceleration "vec a=(hat i+hat j)(m/s^(2))".Find the radius of curvature of the particle at "t=1sec".

A particle has an initial velocity of 4hat(i)+4hat(j) m//s and an acceleration of -0.4 hat(i) m//s^(2) at what time will its speed be 5 m//s?

A particle of mass 2 kg is moving with velocity vec(v)_(0) = (2hat(i)-3hat(j))m//s in free space. Find its velocity 3s after a constant force vec(F)= (3hat(i) + 4hat(j))N starts acting on it.

A particle is moving in x-y plane. At an instant, it has velocity (4 hat (i) + 4 hat(j)) m//s and acceleration (3 hat(i) + 5 hat(j)) m//s^(2) At that instant, the radius of curvature of its path will be :

A charged particle moving in a magnetic field vec(B) = hat(i) - hat(j) tesla, has an acceleration of 2 hat(i) + alpha hat(j) at some instant. Find the value of alpha

Two particles of equal mass have velocities vec v_1 = 2 hat i= m//s^-1 and vec v_2 = 2hat j m//s^-1 . First particle has an acceleration vec a_1= (3 hat i+ 3 hat j) ms^-2 while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of.

The velocity of a particle of mass 2 kg is given by vec(v)=at hat(i)+bt^(2)hat(j) . Find the force acting on the particle.

A particle of mass 2kg moves with an initial velocity of (4hat(i)+2hat(j))ms^(-1) on the x-y plane. A force vec(F)=(2hat(i)-8hat(j))N acts on the particle. The initial position of the particle is (2m,3m). Then for y=3m ,

If vec a=hat i-hat j and vec b=-hat j+2hat k, find (vec a-2vec b)*(vec a+vec b)

Velocity of a particle at some instant is v=(3hat i + 4hat j + 5hat k) m//s . Find speed of the particle at this instant.