Question

a particle moves along the x-axis in such a way that its position at time t is given by
x=3t^4-16t^3+24t^2 for -5≦t≦5
determine the velocity and acceleration of the particle at time t. (I know you find the 1st or 2nd derivative i just don't know what to do after that)

Answer 1

did you get the derivatives?

Answer 2

you are finding

\[\frac{dx}{dt}....and.....\frac{d^2x}{dt^2}\]

Answer 3

I got the derivatives but is that all you need? wouldn't you have to find a "t" value for each?

Answer 4

the velocity and acceleration at some time "t" are still functions of t. It could have asked you to find velocity at t=3, for example, but it didn't... just find velocity at any time "t". Although remember that t is limited to being between -5 and 5.

Answer 5

well the velocity is the 1st derivative, and you have that in terms of t.

the 2nd derivative is the acceleration ,,, and again that is written in terms of t.

Answer 6

so I think you have completed the question.

Answer 7

So would it be 12t^3-48t^2+48t= 0 and you get t=2, t=0 for the t value?

Answer 8

for the first derivative anyway

Answer 9

well that is solving the 1st derivative for t, which gives the stationary points, or where the velocity is equal to zero.

Answer 10

right so does zero count as time "t" because it is in between 5 and -5

Answer 11

well its interesting that your conditions allow t to be negative... I would have thought that it was displacement that is limited to [-5, 5]

Answer 12

and t = 0 is the initial condition... so the particle initially stated from rest

Answer 13

oops started