A particle has a constant acceleration of a = (5.5 m/s2)i + (3.7 m/s2)j . At time t = 0, the velocity is zero and the position vector is r0 = (9.8 m)i.
(a) Find the velocity and position vectors as functions of time t.
(b) Find the equation of the particle's path in the xy plane.
(a) Well, well, we have a particle with some fancy acceleration. Let's get to work, shall we?
To find the velocity as a function of time, we can integrate the acceleration with respect to time. So, let's integrate that bad boy:
v(t) = ∫ a dt
= ∫ (5.5 m/s^2)i + (3.7 m/s^2)j dt
= (5.5 m/s^2)t i + (3.7 m/s^2)t j + C
Now, C is just a constant of integration. Since at t = 0, the velocity is zero, we have:
v(0) = (5.5 m/s^2)(0) i + (3.7 m/s^2)(0) j + C
= C
So, C = 0, and we can simplify our velocity function:
v(t) = (5.5 m/s^2)t i + (3.7 m/s^2)t j
Next, to find the position as a function of time, we integrate the velocity:
r(t) = ∫ v dt
= ∫ [(5.5 m/s^2)t i + (3.7 m/s^2)t j] dt
= (2.75 m/s^2)t^2 i + (1.85 m/s^2)t^2 j + C1
Again, C1 is just another constant of integration. With the initial position r0 given as (9.8 m)i at t = 0:
r(0) = (2.75 m/s^2)(0)^2 i + (1.85 m/s^2)(0)^2 j + C1
= 9.8 m i + C1
So, C1 = 9.8 m, and we have our final position vector function:
r(t) = (2.75 m/s^2)t^2 i + (1.85 m/s^2)t^2 j + 9.8 m i
(b) Now, for the equation of the particle's path in the xy plane. Since there's no acceleration in the z-direction, the particle will remain in the xy plane. Therefore, the path is just the trajectory of the position vector r(t) projected onto the xy plane.
That means we can ignore the j-component of the position vector, giving us the equation of the path in the xy plane:
x(t) = (2.75 m/s^2)t^2
y(t) = 9.8 m
So, the equation of the particle's path in the xy plane is y = 9.8 m. Not very exciting, I must say. But hey, at least we solved it with a touch of clownish expertise, don't you agree?
To find the velocity and position vectors as functions of time, we can integrate the acceleration vector.
(a) The acceleration vector is given as a = (5.5 m/s^2)i + (3.7 m/s^2)j.
Integrating the acceleration vector with respect to time, we can find the velocity vector:
v(t) = ∫a dt
For the x-component of the velocity vector:
∫(5.5 m/s^2) dt = 5.5t + C1
For the y- component of the velocity vector:
∫(3.7 m/s^2) dt = 3.7t + C2
Since the velocity at time t=0 is zero, we have:
v(0) = (5.5 * 0 + C1)i + (3.7 * 0 + C2)j = 0
This implies that C1 = 0 and C2 = 0.
Therefore, the velocity vector as a function of time is:
v(t) = (5.5t)i + (3.7t)j
Now, let's find the position vector.
Integrating the velocity vector with respect to time, we can find the position vector:
r(t) = ∫v dt
For the x-component of the position vector:
∫(5.5t) dt = (5.5/2)t^2 + C3
For the y-component of the position vector:
∫(3.7t) dt = (3.7/2)t^2 + C4
Given that the initial position vector r(0) = (9.8 m)i, we have:
r(0) = ((5.5/2) * 0^2 + C3)i + ((3.7/2) * 0^2 + C4)j = 9.8i
This implies that C3 = 9.8 and C4 = 0.
Therefore, the position vector as a function of time is:
r(t) = ((5.5/2)t^2 + 9.8)i + ((3.7/2)t^2)j
(b) To find the equation of the particle's path in the xy plane, we eliminate the time parameter in the position vector.
From the position vector, we have:
x(t) = (5.5/2)t^2 + 9.8
y(t) = (3.7/2)t^2
From these equations, we can express t in terms of x and y:
t = sqrt((2(x - 9.8))/(5.5))
t = sqrt((2y)/(3.7))
By equating the two expressions for t, we get the equation of the particle's path in the xy plane:
(sqrt((2(x - 9.8))/(5.5))) = sqrt((2y)/(3.7))