Question

At t=0 s, a particle is observed to have position vector Ro= (-3.5,4.0) m. and velocity vector Vo= (21,12.3) m/s. The particle’s acceleration is constant and has been determined to be a= (2.1,5.4) m/s^2.

a)Determine the particle’s velocity (in Cartesian vector form- (Vx,Vy)) at T= 10.5s.

b)What is the particle’s position (in Cartesian vector form- (x,y)) at T=10.5 s ?

Answer 1

A) From a motion with constant accleration, the velocity as a function of time is:

$\vec v = \vec v_0 + \vec a t$

$\vec v = (21,12.3)m/s + (2.1,5.4)m/s^2(10.5 s)$

$\vec v = (43.05,69)m/s$

B) The equation for the position as a function of time:

$\vec R =\vec R_0 + \vec v_0t + \vec a \frac{t^2}{2}$

$\vec R =(-3.5,4.0)m + (21,12.3)m/s(10.5 s)+(2.1,5.4)m/s^2\frac{(10.5 s)^2}{2}$

$\vec R =(332.8,430.8) m$