Question

An object of mass 3kg is at rest.Now a force Fvector=6t2icap+4tjcap is applied on the object then velocity of object at t=3s is

Answer 1

F=m∗a

a=Fm

a=2t2i+2t3j

a=dvdt

dvdt=2t2i+0.66

tj

given initially at rest

∫v0dv=∫30(2t2i+0.66tj)dt

v=[1.33t3i+0.33t2]30

v=17.82i+2.97j

Taking resultant

v=17.822+2.972−−−−−−−−−−−√

v=18.065m/s