This seems easy but I still need help?!! Also how do you find the acceleration?

Question

The motion of a particle on the x-axis is governed by the given equation: x= -11 + 12t + 3t² + 4t³ - t^4. Find the location of the particle and the velocity when the acceleration is a maximum.

Answer 1

x = -11 + 12t + 3t² + 4t³ - t^4
v = 12 + 6t + 12t² - 4t³
a = 6 + 24t - 12t²
for max or min a,
24 - 24t = 0
t = 1 (and this yields a miximum)
a = 6 + 24 - 12 = 18 units/sec²
v = 12 + 6 + 12 - 4 = 26 units/sec
x = -11 + 12 + 3 + 4 - 1 = 7 units

Answer 2

Acceleration = d^2y/dx^2 = 6 + 24t - 12t^2

Acceleration is maximum when its derivative is zero.

ie; 24 - 24t = 0. giving t = 1

Now location is given by x = -11 + 12t + 3t^2 +4t^3 - t^4

So at t = 1:

x = -11 + 12 + 3 + 4 - 1 = 7

The velocity is given by dx/dt = 12 + 6t + 12t^2 - 4t^3

so at t = 1:

velocity = 12 + 6 + 12 - 4 = 26

Hope that helps..

Answer 3

Velocity is the change in x over the change in time. In your case, dx/dt, therefore take the first derivative:

Velocity = dx/dt = 12 + 6t + 12t^2 - 4t^3

Acceleration is the change in velocity at a given time, or the second derivative, so take the derivative again:

Acceleration = d2x/dt2 = 6 + 24t - 12t^2

Acceleration is at a maximum when

d3x/dt3= 24-24t = 0 (review local maxima and minima)

As the maximum is at the value of 0 for the derivative of the acceleration.

Solve for t and plug into the equation for location and velocity.

Answer 4

x= -11 + 12t + 3t² + 4t³ - t^4
v = dx/dt = 12 + 6t + 12t^2 - 4t^3
a = dv/dt = 6 + 24t - 12t^2; solve for t when a = 0.

Put your t answers (there are two) into the x and v equations to solve for location and velocity at maximum acceleration.

Answer 5

this seems easy...so You must be pretty intelligent in physics. I would combine all like terms and I'm guessing t= time. combine your likes terms and solve for x. you can do that of course. but when you mention acceleration this is where you need to know about gravity, is the particle moving toward the earth? if so in a vertical free fall ,after one second is 32 ft./sec. After 2 sec. your velocity is 64 ft/ sec. after 3 seconds your at 96/second and so on. the max ecceleration depends on the distance from the start of the fall and the point of ground zero in ft.

Answer 6

acceleration = the second derivative.

To find the maximum of the acceleration, differentiate a third time and find the x that makes this equal to zero.

Plug this x into the original function to get its location; and into the first derivative for the velocity.