Question

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 !

Answer 1

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
v(1)i=-2 m/s

check that.

Answer 2

0i come from where ?

Answer 3

Also,9-2-4=3 !!

Answer 4

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.