the velocity of a particle is given by v=[16t^2i+4t^3j +(5t+2)k]m/s, where t is in seconds. If the particle is at the origin when t=0, determine the magnitude of the particle's acceleration when t=2s. What is the x,y,z coordinate position of the particle at this instant.

The acceleration is

dv/dt = 32t i + 12 t^2 j + 5 k
Plug in t = 2 to get the acceleration vector at t =2.

The position vector is the integral of v(t),
r(t) = [(16/3)t^3 + a]i +
(t^4 + b)j + [(5/2)t^2 + 2t + c] k
where a, b and c are constants of integration. Use the position at t = 0 to solve for them. It appears that a=b=c=0

Well, well, well, looks like we've got ourselves a physics problem here. Let's dive right in!

To find the magnitude of acceleration, we need to differentiate the given velocity equation with respect to time. So, hold onto your hats, here comes some calculus!

Taking the derivative of the x-component of the velocity equation, we get:

vx = 16t^2

Differentiating again with respect to time, we find:

ax = 32t

Doing the same for the y-component, we have:

vy = 4t^3

Differentiating two times, we'll arrive at:

ay = 24t^2

Now for the z-component:

vz = (5t + 2)

Taking two derivatives, we have:

az = 0

Now, to find the magnitude of acceleration, we use the formula:

|a| = √(ax^2 + ay^2 + az^2)

Plugging in the values, we get:

|a| = √((32t)^2 + (24t^2)^2 + 0^2)

Simplifying this, we get:

|a| = √(1024t^2 + 576t^4)

Now, when t = 2s, we can substitute it into the expression for |a| to find the magnitude of acceleration:

|a| = √(1024(2)^2 + 576(2)^4)

Calculating this, we find:

|a| ≈ 407.76 m/s^2

Now, to determine the position of the particle at t = 2s, we integrate the individual components of the velocity equation with respect to time.

Integrating the x-component, we have:

x = ∫(16t^2) dt = 16∫t^2 dt = 16(t^3/3) = (16/3)t^3

Integrating the y-component, we get:

y = ∫(4t^3) dt = 4∫t^3 dt = (4/4)t^4 = t^4

And integrating the z-component:

z = ∫((5t + 2) dt) = (5/2)t^2 + 2t

Now, plugging in t = 2s, we find:

x = (16/3)(2)^3 = 128/3

y = (2)^4 = 16

z = (5/2)(2)^2 + 2(2) = 12

So, the x, y, z coordinates of the particle when t = 2s are approximately (42.67, 16, 12).

Hope that answers your question with a splash of humor!

To find the magnitude of the particle's acceleration at t=2s, we need to take the derivative of the velocity equation with respect to time. The acceleration is given by the second derivative of the position equation.

Given:
v = 16t^2i + 4t^3j + (5t+2)k

Taking the derivative of v with respect to t, we get:
a = dv/dt = d/dt(16t^2i + 4t^3j + (5t+2)k)
= 32ti + 12t^2j + 5k

Plugging in t=2s, we have:
a = 32(2)i + 12(2^2)j + 5k
= 64i + 48j + 5k

Now, let's calculate the magnitude of the acceleration:
|a| = sqrt((64)^2 + (48)^2 + (5)^2)
= sqrt(4096 + 2304 + 25)
= sqrt(6425)
≈ 80.25 m/s^2

Thus, the magnitude of the particle's acceleration when t=2s is approximately 80.25 m/s^2.

To find the x, y, z coordinates of the particle at t=2s, we integrate the velocity function with respect to time.
The position equation is given by the integral of v with respect to t:

r = ∫v dt = ∫(16t^2i + 4t^3j + (5t+2)k) dt
= (16/3)t^3i + (t^4/2)j + ((5t^2)/2 + 2t)k

Now, plugging in t=2s:
r = (16/3)(2^3)i + (2^4/2)j + ((5(2^2))/2 + 2(2))k
= (128/3)i + 8j + (20 + 4)k
= (128/3)i + 8j + 24k

Therefore, at t=2s, the x, y, z coordinate position of the particle is approximately (128/3, 8, 24).

To find the magnitude of the particle's acceleration when t=2s, we need to differentiate the velocity vector with respect to time (t) to get the particle's acceleration vector.

Given velocity vector v = [16t^2i + 4t^3j + (5t+2)k] m/s,

Acceleration vector a = dv/dt = d/dt [16t^2i + 4t^3j + (5t+2)k]

Differentiating each component of the velocity vector separately, we get:

a = [d(16t^2)/dt i + d(4t^3)/dt j + d((5t+2))/dt k]
a = [32t i + 12t^2 j + 5 k]

Now, we substitute t=2s into the acceleration vector to find its magnitude:

a(2) = [32(2) i + 12(2)^2 j + 5 k]
a(2) = [64 i + 48 j + 5 k]

The magnitude of the acceleration vector is found using the formula:

Magnitude of a = √(a_x^2 + a_y^2 + a_z^2)

Magnitude of a = √(64^2 + 48^2 + 5^2)
Magnitude of a = √(4096 + 2304 + 25)
Magnitude of a = √(6425)
Magnitude of a ≈ 80.13 m/s^2

To find the x, y, z coordinate position of the particle at t=2s, we need to integrate the velocity vector with respect to time.

Given velocity vector v = [16t^2i + 4t^3j + (5t+2)k] m/s,

Integrating each component of the velocity vector separately, we get:

x = ∫(16t^2) dt = (16/3)t^3 + C1
y = ∫(4t^3) dt = (4/4)t^4 + C2
z = ∫((5t+2)) dt = (5/2)t^2 + 2t + C3

Applying the initial condition, when t=0, the particle is at the origin (x=0, y=0, z=0). Therefore, C1 = C2 = C3 = 0.

Substituting t=2s into the position equations, we get:

x(2) = (16/3)(2)^3 = (16/3)(8) = 128/3 m
y(2) = (4/4)(2)^4 = (4/4)(16) = 16 m
z(2) = (5/2)(2)^2 + 2(2) = (5/2)(4) + 4 = 10 + 4 = 14 m

Therefore, the x, y, z coordinate position of the particle at t=2s is (128/3, 16, 14).