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.