At t=0, a particle leaves the origin with a velocity 6m/s in the positive x direction and moves in the x-y plane with a constant acceleration (-2i-4j)m/s2
How far the particle from the origin at t=1 ?
displacement=vi*t+1/2 a t^2
= 6i*1+1/2 (-2i-4j)*1
combine terms. Then how far is = sqrt (icomponent^2+jcomponent^2)
The only issue, is x direction the i direction?
To find the distance of the particle from the origin at t=1, we first need to find the position vector of the particle at t=1.
Given:
Initial velocity, v0 = 6m/s
Acceleration, a = -2i - 4j m/s^2
We know that the position vector can be expressed as the sum of the initial position vector and the displacement vector. The displacement vector can be calculated as the product of time and the velocity vector, plus half the product of time squared and the acceleration vector.
Let's calculate the displacement vector at t=1:
Displacement vector, Δr = (v0 * t) + (0.5 * a * t^2)
Substituting the given values, we have:
Δr = (6m/s * 1s) + (0.5 * (-2i - 4j) * (1s)^2)
Simplifying further, we get:
Δr = 6m/s - (i + 2j) * 0.5m/s^2
Δr = 6i - i - 2j
Δr = 5i - 2j
Therefore, at t=1, the particle is located 5 units in the positive x-direction and 2 units in the negative y-direction from the origin.
To find the distance of the particle from the origin at t=1, we can use the equations of motion.
Given:
- Initial velocity (u) = 6 m/s (in the positive x direction)
- Acceleration (a) = -2i - 4j m/s^2
- Time (t) = 1 second
Step 1: Find the final velocity (v) using the equation v = u + at
Since the particle is moving in the x-y plane, the final velocity will be given by:
v = v_xi + v_yj
where v_x and v_y are the components of velocity in the x and y directions, respectively.
v_x = u_x + a_xt
= 6 + (-2)(1)
= 4 m/s
v_y = u_y + a_yt
= 0 + (-4)(1)
= -4 m/s
So, v = 4i - 4j m/s
Step 2: Find the displacement using the equation s = ut + (1/2)at^2
Since the particle is moving in the x-y plane, the displacement will be given by:
s = s_xi + s_yj
where s_x and s_y are the components of displacement in the x and y directions, respectively.
s_x = u_xt + (1/2)a_xt^2
= 6(1) + (1/2)(-2)(1)^2
= 4 m
s_y = u_yt + (1/2)a_yt^2
= 0 + (1/2)(-4)(1)^2
= -2 m
So, s = 4i -2j
Step 3: Find the magnitude of the displacement to get the distance from the origin.
The magnitude of the displacement is given by:
distance = |s| = √(s_x^2 + s_y^2)
distance = √(4^2 + (-2)^2)
= √(16 + 4)
= √20
= 2√5 m
Therefore, the particle is √20 meters away from the origin at t=1 second.