differentiation question...?

Question

the displacement x m of a particle at time t seconds is given by x=(ke^(-3t))(2t^2+t), where k is a constant. if the initial velocity of the particle is 3m/s, determine the initial acceleration. find also the maximum displacement of the particle.

pls show the steps

Answer 1

First determine the acceleration function by differentiating twice.
v=dx/dt=
(-3ke^(-3t))*(2t^2 +t)+(ke^(-3t))(4t+1)
=(ke^(-3t))*(-6t^2 +t+1)

Solve for k via t=0
3 m/s = (k)

a=dv/dt=
(-3ke^(-3t))(-6t^2 +t+1)+(ke^(-3t))(-12t +1)
=(ke^(-3t))*(18t^2-15t-2)

Then a(t=0)
=-2k
a(0)= -6 m/s

You can find maximum displacement by
setting dx/dt=0
(ke^(-3t))*(-6t^2 +t+1)=0
-6t^2 + t + 1=0
(3t+1)(-2t+1)
Therefore maximum/minimum displacement occurs at
t= -1/3
and
t=1/2
x(-1/3)=(ke^(1))((2/9)-(1/3))
x(-1/3)= 3*e*(-1/3)
x(-1/3)= -e
x(1/2)=(ke^(-3/2))(1)
x(1/2)= 3*e^(-1.5)

So the maximum displacement corresponds to t=(1/2)
and minimum corresponds to t= -1/3

Answer 2

This is a Cauchy problem....
You have first to derive the equation:

x'(t)=(-3ke^(-3t))*(2t^2+t)+
(4t+1)*(ke^(-3t))

You also know that v(0)=x'(0)=3m/s. You have to put this into the equation:

x'(0)=(-3ke^(0))*(0)+(1)*
(ke^(0))=3
x'(0)=k=3

You have found the constant. The equation and its derivate become

x(t)=(3e^(-3t))*(2t^2+t)

x'(t)=(-9e^(-3t))*(2t^2+t)+
(4t+1)*(3e^(-3t))

For the acceleration you have to derivate once again:
a(t)=v'(t)=x''(t)

x''(t)=(27e^(-3t))*(2t^2+t)+
(4t+1)(-9e^(-3t))+
(4t+1)*(-9e^(-3t))+12e^(-3t)

x''(0)=27*0+1*(-9)+1*(-9)+12=
-6m/s^2

For the maximum displacement you have to consider the zero of the first derivate

0=(-9e^(-3t))*(2t^2+t)+
(4t+1)*(3e^(-3t))

You can put -3e^(-3t) in evidence, and obtain

-3e^(-3t)*(6t^2-7t-1)=0

This is verified if at least one of the two terms is null.
The first is a negative exponential, so the solution is t=PLUS INFINITE.
The second is solved as an usual second-degree equation, and the solutions are t=1/2; t=-1/3. This last is unacceptable (i suppose).
Maximum x-desplacement is

x(1/2)=0.22313016

...well, i suggest you to control.
And, please take it easy, not to see Yahoo Answers as an alternative to study (i'm not judging, i am only warning, perhaps you don't even need; in case i beg your pardon).

Let me know if it's been useful.

PS I have had to write some expressions on more than one line because i could not not see them entirely in the preview.

Answer 3

dx/dt =k( -3e^-3t(2t^2+t)+e^-3t(4t+1))at t=0
v= k(1) = 3 so k =3
dx/dt=0 -3(2t^2+t)+4t+1=0
6t^2-t-1=0 t= (1+5)/12 =1/2 s
xmax = 3e^-3/2 m

3*e^-3t)(-6t^2+t+1) = velocity
acceleration = 3[-3e^-3t(-6t^2+t+1)+e^-3t(-12t+1)]
At t=0 accel= 3(-3+1) = -6m/s^2