Deriving v and a from a x(t)-function

1. Sep 1, 2010

astenroo

1. The problem statement, all variables and given/known data
Hi again folks! I just need confirmation on this, since I'm quite sure I've got it right. Anyway here it goes.

The position of a particle is given by the function x(t)= A + Bt + Ct² + Dt³. The values of the constants are the following: A= 4,0; B=2,0; C= -3 and D=1,0.
a) Give the units of the constants if [x]=m and [t]=s
b) What is the velocity and acceleration at t=3s?
c) Give an expression for the velocity v=v(t) and make a sketch for the time interval 0 to 3s
d) From the graph, give the acceleration at t=1s
e) Compute the acceleration at t=1s by examining the quota delta t / delta v for the time intervals delta t= 1,0s; 0,5s; 0,1s and 0,001s

2. Relevant equations

3. The attempt at a solution
a) since [x]=m and [t]=s, then [A]=m, = ms$$^{-1}$$, [C]= ms$$^{-2}$$ and [D]= ms$$^{-3}$$
b) I differentiate once to get the expression for v, v=dx/dt = 3Dt² + 2 Ct + B which gives the instantaneous velocity (perhaps speed in this case) as 11 m/s. Differentiate a second time to get a, a=d²x/dt²= 6Dt + 2C which gives the acceleration at t=3s to be 12m/s²
c) The expression of v I got to be this (taking the constants into account) v= v(t) = 3t² - 6t + 2. this gives a parabola (or part of one) with its apex at v = -1 at t=1. So, the acceleration would be 0 m/s².
d) With limiting function and the given time intervals a approaches zero as delta t approaches zero

I'd really like confirmation on this one :) Thanks!

2. Sep 1, 2010

rock.freak667

They look correct to me.

3. Sep 1, 2010

Thank you :)