x=40+12t−t3 ∴ Velocity,
v==12−3t2 When particle come to rest,
=v=0 ∴12−3t2=0⇒3t2=12 ⇒t=2sec Distance travelled by the particle before coming to rest
ds=vdt or
s=(12−3t2)dt =12t−|02 s=12×2−8=24−8 =16m (OR) At
t=0, particle is at , let's say
x distance, from
O; then putting
t=0 in the given displacement-time equation we get;
x=40+12(0)−(0)3=40m Particle comes to rest that means velocity of particle becomes zero after travelling certain displacement ; let's say the time be t. then after differentiating the given displacement-time equation w.r.t. time we get velocity - time equation
v=12−3t2 when the particle comes to rest ):
at time
t=t v=0; =>12−3t2=0 =>t=2s Then, at
t=2 s we are at , let's say
x′ distance from
O;
put this value oft
(=2) in given displacement-time equation,
we get;
x′=40+12(2)−(2)3 =56m Further; We have seen that the particle started his journey when it is at
40m from the point
O. And came to rest at
56m from the point
O. then the particle traveled a distance of:
56−40=16m