Math!!! Help!!!!?

Question

A particles moves along the x –axis so that, at any times t>0 its acceleration is given by a(t)=6t +6, At time t=0, the velocity of the particles is -9, and its position is -27.

For what values of t>0 is the particles moving to the right?

Answer 1

a(t)=6t +6,

At time t=0, the velocity of the particles is -9, and its position is -27

a(t)=d^2x/dt^2 = 6t +6 integrate

dx/dt = 3 t^2 + 6 t +c

at t=0 , v = - 9>>>> c = - 9

dx/dt = 3 t^2 + 6 t -9 integrate

x = t^3 + 3 t^2 -9t + c1

at t=o m x= - 27 >>> c1 = -27

x = t^3 + 3 t^2 -9t - 27

the x is posive when

t^3 + 3 t^2 -9t - 27=0

t^2 (t - 3) + 6 t (t - 3)+ 9 ( t - 3)
t = 3, and t^2+ 6 t + 9 =0 >>> (t + 3)^2 =0
t = - 3 and -3 (negative time values discarded)

so after t =3 its moving towards right or +x axis

Answer 2

find the velocity function first, which means to take the antiderivative of the a(t). so v(t)=3t^2+6t+c,(c is any constant).
At time t=0, V(0)=3(0)^2+6(0)+c=-9, then you will get c=-9. finally, you get v(t)=3t^2+6t-9, and just set 3t^2+6t-9>0 then you get -3 and 1. The value for the time must be postive, so the correct answer is t >1. i hope it is not too late to help you!

Answer 3

accelaration is 6t+6, then you can write the distance formula as bellow:

d=att+bt+d0,

that in it:

a=6t+t for t>0

d=(6t+6)tt+bt+d0

for finding the velocity (at t=0) you derive the above equation:

v0=d(6ttt+6tt+bt+d0)/dt (when t=0)=18tt+12t+b=0+b=b
assum that v=-9 in t=0, so b=-9

now for finding d0 you can put t=0 in the original equation,
so d0=-27

now you can rewrite your new equation with the factors a, b and d0:


d=(6t+6)tt-9t-27

d=6ttt+6tt-9t-27

this is your movement equation,
now, moving to the right means, your velocity is right hand and d>0,
so 6ttt+6tt-9t-27>0 and the deraive of the equation should be more than 0,
18tt+12t-9>0 and 6ttt+6tt-9t-27>0, you can fine the values of t that make the two statement positive.(this section belongs to 12 years ago and I hardly remember it)

Answer 4

value of t is 19