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

Explanation

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.