A particle moves long a straight line so that at time $t\geq 0$ its acceleration is given by the function $a(t)=4t$. At time $t=0$, the velocity of the particle is 4 and the position of the particle is 1. Which of the following is an expression for the position of the particle at time $t\geq 0$?

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We can integrate the acceleration equation with respect to time to obtain the velocity equation.

$$v(t) = \int a(t)   dt$$ $$v(t) = \int 4t   dt$$ $$v(t) = 2t^2+C$$

Using our initial condition:

$$v(0)=4$$ $$4 = 2(0)^2+C$$ $$4=C$$ $$v(t)= 2t^2+4$$

Integrate again to obtain the position equation.

$$s(t)=\int v(t)   dt$$ $$s(t) = \int 2t^2+4   dt$$ $$s(t)= \frac{2}{3}t^3+4t+C$$

Using our initial condition:

$$s(0)=1$$ $$1 = \frac{2}{3}(0)^3+4(0)+C$$ $$1 = C$$ $$s(t)=\boxed{\frac{2}{3}t^3+4t+1}$$