At t= 0, a particle starts from rest at x= 0, y= 0, and moves in the xy plane with an acceleration ->a = (4.0ihat+ 3.0jhat)m/s^2. Assume t is in seconds.
Determine the position of the particle as a function of time t.
Express your answer in terms of the unit vectors ihat and jhat.
A = 4 i + 3 j
V = integral A dt = 4 t i + 3 t j + initial velocity which is 0
R = integral V dt = 2 t^2 i + (3/2)t^2 j + initial position which is 0
To determine the position of the particle as a function of time t, we will integrate the given acceleration vector with respect to time twice.
First, let's integrate the x-component of the acceleration:
∫(4.0î) dt = 4.0t + C₁, where C₁ is the constant of integration.
Next, we integrate the y-component of the acceleration:
∫(3.0ĵ) dt = 3.0t + C₂, where C₂ is the constant of integration.
Now, we have the components of the velocity vector:
vₓ = 4.0t + C₁
vᵧ = 3.0t + C₂
To find the position vector, we integrate the velocity components with respect to time:
∫(4.0t + C₁) dt = 2.0t² + C₁t + C₃, where C₃ is the constant of integration for the x-component of the position vector.
∫(3.0t + C₂) dt = 1.5t² + C₂t + C₄, where C₄ is the constant of integration for the y-component of the position vector.
Finally, we have the components of the position vector as functions of time t:
x(t) = 2.0t² + C₁t + C₃
y(t) = 1.5t² + C₂t + C₄
Given that the particle starts from rest at position (0, 0) at t = 0, we can substitute these initial conditions into the position equations:
At t = 0:
x(0) = 2.0(0)² + C₁(0) + C₃ = 0
y(0) = 1.5(0)² + C₂(0) + C₄ = 0
From these equations, we find that C₃ = 0 and C₄ = 0.
Therefore, the position of the particle as a function of time t is:
x(t) = 2.0t² + C₁t
y(t) = 1.5t² + C₂t
Note: The constants C₁ and C₂ can be determined if initial conditions or additional information is provided.
To determine the position of the particle as a function of time, we need to integrate the acceleration twice with respect to time.
1. First, integrate the acceleration ->a with respect to time to find the velocity ->v as follows:
∫->a dt = ∫(4.0ihat + 3.0jhat) dt
The integral of the acceleration with respect to time is:
->v = (4.0t + C1)ihat + (3.0t + C2)jhat
Where C1 and C2 are constants of integration. In this case, the particle starts from rest at t=0, which means the initial velocity (v = 0) at t=0. Therefore, C1 and C2 are both 0, so the equation becomes:
->v = 4.0tihat + 3.0tjhat
2. Now, integrate the velocity ->v with respect to time to find the position ->x as follows:
∫->v dt = ∫(4.0tihat + 3.0tjhat) dt
The integral of the velocity with respect to time is:
->x = (2.0t^2 + C3)ihat + (1.5t^2 + C4)jhat
Where C3 and C4 are constants of integration. In this case, the particle starts from the origin at t=0, which means the initial position (x = 0, y = 0) at t=0. Therefore, C3 and C4 are both 0, so the equation becomes:
->x = 2.0t^2ihat + 1.5t^2jhat
So, the position of the particle as a function of time t is:
->x = 2.0t^2ihat + 1.5t^2jhat