The following question refers to the following expression for a particle's position:
X = 3t^2 - 4t + 3
What is the particle's acceleration after 3 seconds?
I keep getting 1 m/s2 but that's not an answer choice.
I found instantaneous velocity at t=0 (3 m/s) and at t = 3 (6 m/s) and just calculated change in velocity over change in time....
Where am I going wrong?
well, if you have had any calulus
v = dx/dt = 6 t - 4
a = dv/dt = 6 forever
if you have just ha motion under constant acceleration
x = xi + Vi t + (1/2) a t^2
so
(1/2) a = 3
a = 6 again