Plz answer the following Question?

Question

The position of a particle is given by s=4t^3+3t^2+5 Where s is in metres and t in seconds
Determine the time 't; and acceleration 'a' when velocity 'v'=0

Answer 1

some of the answers obtained and assertions on the validity of the answers given by others are wrong. so, here it is:
v = ds/dt = 12t^2 + 6t
v = 0 => t = 0 or -1/2
{t = -1/2 is also a valid solution. It just means 1/2 a unit before u started counting time from 0}
a = dv/dt = 24t +6
when v = 0, t = 0 or -1/2,
so a = 6 or -6

Answer 2

s = 4t^3 + 3t^2 + 5
v = ds/dt = 12t^2 + 6t
a = dv/dt = 24t + 6
@v = 0
t(2 + t) = 0
t = 0, -2 discard -2
t = 0 when v = 0
a(0) = 6 m/s^2

Answer 3

v = ds/dt = 12t^^2+6t =0

t=-1/2

Its a nonsense problem. There is no such time.

If you change the sign of the second term, t = 1/2 sec

A = dv/dt = 24t -6 = 12-6 = 6 m/s^^2

Answer 4

remember that the first derivative of position is velocity.
the second derivative is acceleration.
set the velocity equation (ds/dt) to 0, and you get a quadratic. solve for t.
next, take acceleration (ds/dt/dt), plug in the value of t you got, and solve for a.