posted by meer on .
At t = 0, a particle leaves the origin with a velocity of 9 m/s in the positive y direction and moves in the xy plane with a constant acceleration of (-2 i -4 j) m/s2. At t = 1 s, what is the velocity of the particle in the x direction?
I need explanation for this question !
There is not much explaination to do, it is a vector equation.
using i,j coordinates
v(t)=vi+at=0i+9j +(-2t i)+ (-4t j)
put in t=1, and solve. velocity in the i direction will be let me see
0i come from where ?
i,j are unit vectors, in different directions. Oi means zero in the i direction, as in the initial velocity given.
You cannot 9-2-4 as those were in different directions.
Surely this is in your textbook.