Find the position vector of a particle that has the given acceleration and the specified initial velocity and position.
a(t)= 56t i + sin(t) j + cos(2t) k
v(0)= i
r(0)= j
I need help finding r(t)
Thanks so much for your help!!
v(t) = 28t^2i - cos(t)j + 1/2 sin(2t)k + c
v(0)=0 so c = 1j
v(t) = 28t^2i + (1-cos(t))j + 1/2 sin(2t)k
r(t) = 28/3 t^3i + (t-sin(t))j - 1/4 cos(2t)k + c
r(0)=0 so c = 1/4 k
r(t) = 28/3 t^3i + (t-sin(t))j + 1/4 (1-cos(2t))k
To find the position vector r(t), you need to integrate the given acceleration function twice with respect to time.
Given information:
Acceleration function: a(t) = 56t i + sin(t) j + cos(2t) k
Initial velocity: v(0) = i
Initial position: r(0) = j
Step 1: Integrate the acceleration function to find the velocity function v(t).
Using the acceleration function a(t), integrate each component separately with respect to time:
∫(56t) dt = 28t^2 + C_1
∫(sin(t)) dt = -cos(t) + C_2
∫(cos(2t)) dt = (1/2)sin(2t) + C_3
Now we have the velocity function v(t) as:
v(t) = (28t^2 + C_1) i - cos(t) + C_2 j + (1/2)sin(2t) + C_3 k
To find the constants C_1, C_2, and C_3, we use the given initial velocity:
v(0) = i
Substituting t=0 and equating the components, we get:
C_1 = 0, C_2 = 1, and C_3 = 0
Therefore, the velocity function becomes:
v(t) = 28t^2 i - cos(t) j + (1/2)sin(2t) k
Step 2: Integrate the velocity function to find the position function r(t).
Using the velocity function v(t), integrate each component separately with respect to time:
∫(28t^2) dt = (28/3)t^3 + C_4
∫(-cos(t)) dt = -sin(t) + C_5
∫((1/2)sin(2t)) dt = -(1/4)cos(2t) + C_6
Now we have the position function r(t) as:
r(t) = (28/3)t^3 + C_4 i - sin(t) + C_5 j - (1/4)cos(2t) + C_6 k
To find the constants C_4, C_5, and C_6, we use the given initial position:
r(0) = j
Substituting t=0 and equating the components, we get:
C_4 = 0, C_5 = 1, and C_6 = 0
Therefore, the position function becomes:
r(t) = (28/3)t^3 i - sin(t) j - (1/4)cos(2t) k
So, the position vector of the particle is given by r(t) = (28/3)t^3 i - sin(t) j - (1/4)cos(2t) k.
To find the position vector r(t), we need to integrate the given acceleration a(t) twice with respect to time.
Given:
a(t) = 56t i + sin(t) j + cos(2t) k
v(0) = i
r(0) = j
Let's start by integrating a(t) once to find the velocity vector v(t):
∫[a(t) dt] = ∫[(56t i + sin(t) j + cos(2t) k) dt]
v(t) = ∫[56t i + sin(t) j + cos(2t) k dt]
Integrating each component separately:
∫[56t i dt] = 28t^2 i
∫[sin(t) j dt] = -cos(t) j
∫[cos(2t) k dt] = (1/2)sin(2t) k
So, the velocity vector v(t) becomes:
v(t) = 28t^2 i - cos(t) j + (1/2)sin(2t) k
Now, we integrate v(t) again to find the position vector r(t):
∫[v(t) dt] = ∫[(28t^2 i - cos(t) j + (1/2)sin(2t) k) dt]
r(t) = ∫[28t^2 i - cos(t) j + (1/2)sin(2t) k dt]
Integrating each component separately:
∫[28t^2 i dt] = (28/3)t^3 i
∫[-cos(t) j dt] = -sin(t) j
∫[(1/2)sin(2t) k dt] = -(1/4)cos(2t) k
So, the position vector r(t) becomes:
r(t) = (28/3)t^3 i - sin(t) j - (1/4)cos(2t) k
Therefore, the position vector of the particle at time t is given by:
r(t) = (28/3)t^3 i - sin(t) j - (1/4)cos(2t) k