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 !

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.

To find the velocity of the particle in the x-direction at t = 1s, we can use the equations of motion and basic vector addition.

The given information tells us that the particle starts at the origin (0,0) with an initial velocity of 9 m/s in the positive y-direction (upwards). The constant acceleration of the particle is given as (-2 i -4 j) m/s^2.

The equation of motion that relates displacement (s), initial velocity (u), acceleration (a), and time (t) is given by:
s = ut + 1/2 a t^2

To find the velocity in the x-direction, we need to determine the displacement in the x-direction at t = 1s. Since the initial velocity in the x-direction is 0, the displacement is simply the distance traveled in the x-direction.

To find the displacement in the x-direction, we can use the formula:
s_x = u_x * t + 1/2 * a_x * t^2

Since the initial velocity in the x-direction is 0, u_x = 0. Thus, the equation becomes:
s_x = 0 + 1/2 * a_x * t^2

Now, we need to find the x-component of the acceleration. The given acceleration vector is (-2 i -4 j). The x-component of this vector is -2.

Plugging in the values, we get:
s_x = 1/2 * (-2) * (1)^2

Simplifying the equation, we find:
s_x = -1

Therefore, the displacement in the x-direction at t = 1s is -1 meter.

Since the initial velocity in the x-direction is 0, and there is no acceleration in the x-direction, the velocity in the x-direction will remain 0 throughout the motion. So, at t = 1s, the velocity of the particle in the x-direction is 0 m/s.

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.

Also,9-2-4=3 !!