Question

plz answer it quickly......​
answer is A... explain it

Answer 1

instantaneous acceleration=dv/dt

=(3t^2/3)+2t-3,when t=T

=T^2+2T-3

average acceleration=[v(T)-v(0)]/[T-0]

=[(T^3/3)+T^2-3T]/T

=(T^2/3)+T-3

according to the question

a(inst)=4×a(average)

T^2+2T-3=4×[(T^2/3)+T-3]

T^2+6T-9=0

by solving this we get,

T=-9 and T=3

since negative value is not possible,

T=3 s

answer is option (A)