Question

(12). If x=e^5t, where x is the position of particles in meters , then position of the particles at time t =1 sec is equal to​

Answer 1

$x—>DISPLACEMENT$

$DX/DT—>VELOCITY$

$X=AT³–BT³$

$V=DX/DT=2AT–3BT²$

$A=DV/DT=D²X/DT²$

$2A–2×3BT$

$2A–6BT$

$WHEN A=0 =2A–6BT=0$

$THEREFORE 6BT =2A$

$T=A/3B$