The position vector of a particle is given byvecr= 3 t hati+ 5t^2 hatj + 7 hatk Find the direction in which the particle experiences net force?

The position vector of a particle is given byvecr= 3 t hati+ 5t^2 hatj

1.	The position vector of a particle is given byvecr= 3 t hati+ 5t^2 hatj + 7 hatk Find the direction in which the particle experiences net force?
Answer» SOLUTION :Velocity of the particle, ` VECV = (d vecr)/(DT) = (d)/(dt) (3t) hati + (d)/(dt) (5t^2) hatj + (d)/(dt) hatk ` `(dvecr)/(dt) = 3 hati + 10T hatj ` Acceleration of the particle `vec a = (dvecv)/(dt) = (d^2 vecr)/(dt^2) = 10 hatj` Here, the particle has acceleration only along positive y direction. According to Newtons second LAW, net force must also act along positive y direction. Ir addition, the particle has constant velocity in positive x direction and no velocity in z direction. Hence, there are no net force along x or z-direction.