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:

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