Question

The position of a particle as a function of time is estimated to be: x=(3.00 m/s)t+(1.00m/s^2)t^2-(3.00m/s^3)t^3+(2.00m/s^4)t^4. The particle velocity at the instant when t=2.00s is? The particle acceleration as a function of time is?

Answer 1

acceleration is...\[a=\frac{d^2x(t)}{dt^2}=\frac{dv}{dt}\]where\[v=\frac{dx}{dt}\]