Question

The acceleration of a particle at time t moving along the x-axis is given by: a =4e^2t. At the instant when t=0, the particleis at the point x=2 moving with velocity v=-2. The position of the particle at t=1/2 is?

Answer 1

You have to integrate \(a=4e^{2t}\) twice in order to get the position function.The first integral represents the velocity of the particle. Keep in mind that you have the point \(v(0)=-2\), which will enable you to find the constant of the first integral.

Answer 2

Now, integrate the value you found again to get the the function that represents the position of the article. Use the given point \(x(0)=2\) to find the integration constant.

Answer 3

\[\int\limits 4e ^{2t} dt\]

Answer 4

that sould be v(t)

Answer 5

Exactly!

Answer 6

and then I would have to find C

Answer 7

to do that i plug in -2

Answer 8

You plug t=0 and v=-2.

Answer 9

\[\vec{r''}(t)=\vec{a}(t)\]
\[\vec{r'}(t)=\vec{v}(t)\]
\[\therefore \int\limits_{t_i}^{t_f} \vec{a}(t)dt=\vec v(t)\implies\int\limits_{t_i}^{t_f}\vec v(t) dt=\vec r(t)\]
You're given the inital conditions:
\[\vec{r}(0)=2m; \vec{v}(0)=-2ms^{-1}\]
and
\[\vec{a}(t)=4e^2t\]
Just plug and chug.

Answer 10

Thank You guys