Help Calculus please!!!?

Question

the motion of a particle on the x-axis is governed by the given equation: x=-11+12t+3t^2+4t^3-4t^4. find the location of the particle and the velocity when the acceleration is a maximum.

Answer 1

x = -11 + 12t + 3t^2 + 4t^3 - 4t^4

Taking the 1st derivative gives the velocity.
v = dx/dt = 12 + 6t + 12t^2 - 16t^3

Taking the 2nd derivative gives the acceleration.
a = dv/dt = 6 + 24t - 48t^2

To find when the acceleration is maximum, we have to take the derivative of the acceleration equation and set it equal to zero.

So, da/dt = 24 - 96t = 0.
Thus, t = 1/4.

The location of the particle at this time is then found by substituting t = 1/4 into the location equation.

x = -11 + 12(1/4) + 3(1/4)^2 + 4(1/4)^3 - 4(1/4)^4
= -11 + 3 + 3/16 + 1/16 - 1/64
= -497 / 64

The velocity at this time is found by substituting t = 1/4 into the velocity equation.

v = 12 + 6(1/4) + 12(1/4)^2 - 16(1/4)^3
= 12 + 3/2 + 3/4 - 1/4
= 14

Answer 2

Acceleration is the second derivative.

The zeros of the third derivative are the maxima and minima.

x = -11 + 12t + 3t^2 + 4t^3 - 4t^4
x' = 12 + 6t + 12t^2 -16t^3
x'' = 6 + 24t - 48t^2
x''' = 24 - 96t

24 - 96t =0
96t = 24
t = 24/96
t = 1/4

x = -11 + 12(1/4) + 3(1/4)^2 + 4(1/4)^3 - 4(1/4)^4
x = -11 + 3 + 3/16 + 1/16 - 1/64
x = -7 49/64

Answer 3

You really shoud do your own homework. Take two derivatives to get the acceleration. Then take a derivative of *that* and set it equal to zero to get critical points. Figure out which one(s) give a maximum for the acceleration, and then compute the position and velocity at those points.

Answer 4

x=-11+12t+3t^2+4t^3-4t^4
ds/dt=12+6t+12t^2-16t^3
d^2s/dt^2=6+24t-48t^2
set it to 0 and solve for t
plug in this value of t in the given equation for the location/distance and in ds/dt for velocity

Answer 5

To find where the veloctiy is maximum, set the first derivative of x to zero, and solve for t, then find x at this value of t. To find where the acceleration is zero, set the second derivative of x to zero and solve for t, then find x for this value of t.

Note: When doing these calculations, you have to make sure that you are indeed finding the maximum because the condition for maxima is the same for minima.

Answer 6

supposed t is the time elapsed.

diff w.r.t. t

dX/dt = 12 + 6t + 12t^2 - 16t^3

at stationary point, dX/dt is 0

-16t^3 + 12t^2 + 6t + 12 = 0
t=6.3

to check whether t = 6.3 gives a maximum acceleration,

diff. dX/dt w.r.t. t

d^2X/dt^2 = 6 + 24t -48t^2
d^2X/dt^2 = 6 + 24(6.3) - 48(6.3)^2 < 0, maximum

therefore,
X = -11 +12(6.3) + 3(6.3)^2 + 4(6.3)^3 - 48(6.3)^4
X = -74430.4