HELP please with this question!?

Question

An object moves along a straight line and its position s (in cm) at t seconds is given by s(t) = (t^2 + 3)/t + 2. When is the velocity of the particle equal to 0?

Answer 1

Velocity is the time derivative of displacement. Therefore, take the derivative of the expression for position, s(t), with respect to t, and then set it equal to zero and solve.

s(t) = (t^2 + 3)/t + 2 = t + 3t^-1 + 2
s'(t) = 1 - 3t^-2
s'(t) = 0 ==> 1 - 3t^-2 = 0 ==> 1 = 3/t^2 ==> t^2 = 3 ==> t = sqrt(3)

If s(t) is actually (t^2 + 3) / (t + 2), the differentiation is more difficult.

Answer 2

To simplify, s = t + 3/t +2
v = 0 when ds/dt = 0
Differentiating, ds/dt = 1 - 3/(t^2)

Solving 1 - 3/(t^2) = 0,
we get t^2 = 3.

Hence, t = sqrt 3 or - (sqrt 3)
Since time cannot have a negative value,
the answer is t = sqrt 3.

Answer 3

if you have

s(t) = (t^2+3)/t + 2

that simplifies to

s(t) = t + 3*t^(-1) + 2

take the derivative of position s(t) to get velocity v(t)

v(t) = 1 - 3*t^(-2)

set this to 0.... 0 = 1 - 3*t^(-2)

t = sqrt (3)

Answer 4

s(t) = (t^2 + 3)/t + 2. =t+3/t+2
ds/dt=1-3/t^2.=0
t^2=3
t=sqrt(3)sec.