Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position.
a(t) = 2 i + 6t j + 12t2 k, v(0) = i,
r(0) = 2 j − 7 k
Well, this is a fun physics problem! Let's find the velocity and position vectors step by step:
Step 1: Integrating the acceleration to get the velocity:
Given acceleration a(t) = 2i + 6tj + 12t^2k
To get the velocity v(t), we integrate each component of the acceleration vector with respect to time:
∫2 dt = 2t + C1 (integrating i component)
∫6t dt = 3t^2 + C2 (integrating j component)
∫12t^2 dt = 4t^3 + C3 (integrating k component)
Now we have the general form of the velocity vector:
v(t) = (2t + C1)i + (3t^2 + C2)j + (4t^3 + C3)k
Step 2: Finding the constants using the initial conditions:
Given initial velocity v(0) = i.
Plugging in t = 0 into the general form of the velocity vector gives us:
v(0) = (2(0) + C1)i + (3(0)^2 + C2)j + (4(0)^3 + C3)k
= C1i + C2j + C3k
Since v(0) = i, this means C1 = 1, C2 = 0, and C3 = 0.
Putting these values back into the general form of the velocity vector gives us:
v(t) = (2t + 1)i + (3t^2)j + (4t^3)k
Step 3: Integrating the velocity to get the position vector:
To get the position r(t), we integrate each component of the velocity vector with respect to time:
∫(2t + 1) dt = t^2 + t + C4 (integrating i component)
∫(3t^2) dt = t^3 + C5 (integrating j component)
∫(4t^3) dt = t^4 + C6 (integrating k component)
Now we have the general form of the position vector:
r(t) = (t^2 + t + C4)i + (t^3 + C5)j + (t^4 + C6)k
Step 4: Finding the constants using the initial conditions:
Given initial position r(0) = 2j - 7k.
Plugging in t = 0 into the general form of the position vector gives us:
r(0) = (0^2 + 0 + C4)i + (0^3 + C5)j + (0^4 + C6)k
= C4i + C5j + C6k
Since r(0) = 2j - 7k, this means C4 = 0, C5 = 2, and C6 = -7.
Putting these values back into the general form of the position vector gives us:
r(t) = (t^2 + t)i + (t^3 + 2)j + (t^4 - 7)k
So, the velocity vector is v(t) = (2t + 1)i + (3t^2)j + (4t^3)k, and the position vector is r(t) = (t^2 + t)i + (t^3 + 2)j + (t^4 - 7)k.
Hope that adds some humor to your physics homework!
To find the velocity and position vectors of a particle given the acceleration, initial velocity, and initial position, we can use the equations of motion.
Given:
Acceleration vector, a(t) = 2i + 6tj + 12t^2k
Initial velocity vector, v(0) = i
Initial position vector, r(0) = 2j - 7k
Step 1: Find the velocity vector, v(t).
Since acceleration is the rate of change of velocity, we can integrate the acceleration vector with respect to time to find the velocity vector.
v(t) = ∫a(t) dt
= ∫(2i + 6tj + 12t^2k) dt
Integrating each component separately:
vx(t) = ∫2 dt = 2t + C1
vy(t) = ∫6t dt = 3t^2 + C2
vz(t) = ∫12t^2 dt = 4t^3 + C3
Applying the initial velocity condition v(0) = i:
vx(0) = 0 + C1 = 1
Therefore, C1 = 1
So, the velocity vector v(t) is:
v(t) = (2t + 1)i + (3t^2 + C2)j + (4t^3 + C3)k
Step 2: Find the position vector, r(t).
Since velocity is the rate of change of position, we can integrate the velocity vector with respect to time to find the position vector.
r(t) = ∫v(t) dt
= ∫((2t + 1)i + (3t^2 + C2)j + (4t^3 + C3)k) dt
Integrating each component separately:
rx(t) = ∫(2t + 1) dt = t^2 + t + C4
ry(t) = ∫(3t^2 + C2) dt = t^3 + C2t + C5
rz(t) = ∫(4t^3 + C3) dt = t^4 + C3t + C6
Applying the initial position condition r(0) = 2j - 7k:
rx(0) = 0 + 0 + C4 = 0
Therefore, C4 = 0
ry(0) = 0 + 0 + C5 = 2
Therefore, C5 = 2
rz(0) = 0 + 0 + C6 = -7
Therefore, C6 = -7
So, the position vector r(t) is:
r(t) = (t^2 + t)i + (t^3 + C2t + 2)j + (t^4 + C3t - 7)k
To find the velocity and position vectors of a particle given the acceleration, initial velocity, and initial position, we can use the following equations:
1. Velocity vector (v(t)):
v(t) = ∫a(t) dt + v(0)
2. Position vector (r(t)):
r(t) = ∫v(t) dt + r(0)
Now let's calculate the velocity and position vectors step by step.
1. Velocity vector (v(t)):
Given: a(t) = 2i + 6tj + 12t^2k, v(0) = i
Integrating each component of the acceleration vector:
∫ax(t) dt = ∫2 dt = 2t + Cx
∫ay(t) dt = ∫6t dt = 3t^2 + Cy
∫az(t) dt = ∫12t^2 dt = 4t^3 + Cz
Adding the constant terms:
v(t) = (2t + Cx)i + (3t^2 + Cy)j + (4t^3 + Cz)k
Now, we need to find the value of Cx from the given initial velocity:
v(0) = i, so comparing the i-components:
2(0) + Cx = 1
Cx = 1
Therefore, the velocity vector is:
v(t) = (2t + 1)i + (3t^2 + Cy)j + (4t^3 + Cz)k
2. Position vector (r(t)):
Given: r(0) = 2j - 7k
Integrating each component of the velocity vector:
∫vx(t) dt = ∫(2t + 1) dt = t^2 + t + Dx
∫vy(t) dt = ∫(3t^2 + Cy) dt = t^3 + Cy*t + Dy
∫vz(t) dt = ∫(4t^3 + Cz) dt = t^4 + Cz*t + Dz
Adding the constant terms:
r(t) = (t^2 + t + Dx)i + (t^3 + Cy*t + Dy)j + (t^4 + Cz*t + Dz)k
Now, we need to find the values of Dx, Dy, and Dz from the given initial position:
r(0) = 2j - 7k, so comparing the j and k-components:
0 + Dy = 2 => Dy = 2
0 + Dz = -7 => Dz = -7
Therefore, the position vector is:
r(t) = (t^2 + t + Dx)i + (t^3 + Cy*t + 2)j + (t^4 + Cz*t - 7)k
Note: The values of the constants Cx, Cy, Dx, and Dz can be determined by considering any constraints or additional information given in the problem statement.
a = 2i + 6tj + 12t^2k
v = 2ti + 3t^2j + 4t^3k + c
at t=0, v=i, so c = i and
v = (2t+1)i + 3t^2j + 4t^3k
r = (t^2+t)i + t^3j + t^4k + c
use r(0) to see that c = 2j-7k so
r = (t^2+t)i + (t^3+2)j + (t^4-7)k