Question

Kinematics Exercises[1] The acceleration of a particle in 1-D varies with time as: a(t) = pt² −qt³.Assuming that its initial position and velocity are zero:Find an expressions for the velocity v(t) and the position x(t).

Answer 1

From the acceleration yo can find the rest of the expressions by integrating the initial equations

`\begin{gathered} a(t)=pt^2-qt^3 \\ v(t)=\int a(t)dt=(pt^3)/(3)-(qt^4)/(4)+c \\ d(t)=\int\int a(t)dtdt=(pt^4)/(12)-(qt^5)/(20)+ct+b \end{gathered}`

Is important to put the constants when you integrate, they can affect the result when you solve the system

`\begin{gathered} v(0)=0=(p(0)^3)/(3)-(q(0)^4)/(4)+c \\ 0=c \\ d(0)=(p(0)^4)/(12)-(q(0)^5)/(20)+b=0 \\ 0=b \end{gathered}`

After having the expressions, you replace the assumptions, in this case, v and d are 0, when time t is 0