Question

Position of a particle is given by X equal to 4 + 8t - 4t square metre and find out magnitude of net displacement and total distance covered by the particle from 0 to equal to 2 second​

Answer 1

Solution is in the attachment

Answer 2

x=4+3t-4t^2

x^2=4+3×4-4×4^2

x=4-12+12

=4

v=dx/dt

=8-8t =0

t=1

x (t=1)=8-8×1

d=4+4

=8m