Question

At t= 0, a particle moving in the xy plane with constant acceleration has a velocity of Vi= (3.00i -2.00j) m/s and is at the origin. At t= 3.00 s, the particle's velocity is V = (9.00i + 7.00j) m/s . Find (a) the acceleration of the particle and (b) its coordinates at any time t.

Answer 1

Answer:

(a) a = (2i + 4.5j) m/s^2

(b) r = ro + vot + (1/2)at^2

Explanation:

(a) The acceleration of the particle is given by:

$\vec{a}=\frac{\vec{v}-\vec{v_o}}{t}$

vo: initial velocity = (3.00i -2.00j) m/s

v: final velocity = (9.00i + 7.00j) m/s

t = 3s

by replacing the values of the vectors and time you obtain:

$\vec{a}=\frac{1}{3s}[(9.00-3.00)\hat{i}+(7.00-(-2.00))\hat{j}]\\\\\vec{a}=(2\hat{i}+4.5\hat{j})m/s^2$

(b) The position vector is given by:

$\vec{r}=\vec{r_o}+\vec{v_o}t+\frac{1}{2}\vec{a}t^2$

where vo = (3.00i -2.00j) m/s and a = (2.00i + 4.50j)m/s^2