To find acceleration at time t, we have to differentiate the position vector twice.  

Differentiating the first time gives the velocity:

v(t) = r'(t) = 12t³i + 12tj

Differentiating a second time gives the accelaration: 

a(t) = r''(t) = 36t²i + 12j

Plug in t=1 to solve for the final answer:

a(1) = r''(1) = 36i + 12j

The instantaneous acceleration is found by taking the 2nd derivative of the function and applying thereto the desired variable parameter. First let us calculate the 1st derivative:

f(t) = 4t² * ln(t) will require us to apply the product rule; therefore:

f'(t) = 8t * ln(t) + 4t² * (1/t), which reduces to: 8t*ln(t) + 4t

Taking the second derivative (applying the product rule to the first value), we get:

f''(t) = 8 * ln(t) + 8t * (1/t) + 4, which simplifies to: 8 * ln(t) + 8 + 4 = 8 * ln(t) + 12

The instantaneous acceleration at t = 3.5 is found by solving:

f''(3.5) = 8 * ln(3.5) + 12 = 22.02210374796294 (approx.) or 22.02

The instantaneous acceleration is represented by the second derivative of the positional equation. Let's first calculate the velocity then the acceleration:

v(t) = s'(t) = 15t² – 48t

a(t) = v'(t) = s''(t) = 30t – 48

a(4) = 30 * 4 – 48 = 72

The instantaneous acceleration is represented by the second derivative of the positional equation. Let's first calculate the velocity then the acceleration:

s(t) = 5sin(4t)+ cos(2t)?

v(t) = s'(t) = 20cos(4t) – 2sin(2t)

a(t) = v'(t) = s''(t) = –80sin(4t) – 4cos(2t)

a(π) = –80sin(4π) – 4cos(2π) = –80sin(0) - 4cos(0) = 0 – 4 = –4

The instantaneous acceleration is represented by the second derivative of the positional equation.  Let's first calculate the velocity then the acceleration:

s(t) = 5sin(t/2 + π)?

v(t) = s'(t) = (5/2)cos(t/2 + π)

a(t) = v'(t) = s''(t) = –(5/4)sin(t/2 + π)

We know that sin(x) = –sin(x + π); therefore, sin(t/2 + π) = –sin(t/2). Let's rewrite a(t) as (5/4)sin(t/2).

The acceleration is therefore a(π/2) = (5/4)sin(π/2/2) = (5/4)sin(π/4) = 5/4 * (1/√2) = 4/(4√2) = 5√2/8

Acceleration is given by the second derivative of position.  So for this equation, the second derivative is .  The acceleration is constant through time, so the answer is simply 8.

Since jerk is the rate of change of acceleration, and acceleration is the second derivative of position, the answer is to find the third derivative of position.  Use the chain rule each time.

The acceleration of a particle is given by the second derivative of the position function.  We are given the position function as

 .  

The first derivative (the velocity) is given as

 .  

The second derivative (the acceleration) is the derivative of the velocity function.  This is given as

 .  

Evaluating this at  gives us the answer.  Doing this we get

 .

We know that  and  so the acceleration components are:

Plugging these into the formula for the acceleration and again recognizing that , we get , or, after the square root, .

The acceleration is the derivative of the velocity function.  For the sinusoid component, the chain rule must be used to get the derivative.