In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function. Then find the derivative of the velocity function to get the acceleration function: $\displaystyle p''(t)= v'(t) = a(t)$

In this case, the position function is: $\displaystyle p(t) = 10t^6 - 6t^4 - 5t^2 - 1000t$

Then take the derivative of the position function to get the velocity function: $\displaystyle v(t) = 60t^5 - 24t^3 - 10t - 1000$

Then take the derivative of the velocity function to get the acceleration function: $\displaystyle a(t) = 300t^4 - 72t^2 - 10$

Then, plug $\displaystyle t=1$

 into the acceleration function: $\displaystyle a(1) = 300({\color{Blue} 1})^4 - 72({\color{Blue} 1})^2 - 10$

Therefore, the answer is: $\displaystyle 218$

In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function. Then find the derivative of the velocity function to get the acceleration function: $\displaystyle p''(t)= v'(t) = a(t)$

In this case, the position function is: $\displaystyle p(t) = \sin(3t)$

Then take the derivative of the position function to get the velocity function: $\displaystyle v(t) = 3\cos(3t)$

Then take the derivative of the velocity function to get the acceleration function: $\displaystyle a(t) = -9\sin(3t)$

Then, plug $\displaystyle t=0$ into the acceleration function: $\displaystyle a(0) = -9\sin(3({\color{Blue} 0}))$

Therefore, the answer is: $\displaystyle 0$

In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function. Then find the derivative of the velocity function to get the acceleration function: $\displaystyle p''(t)= v'(t) = a(t)$

In this case, the position function is: $\displaystyle p(t) = \frac{e^{3t}}{3}$

Then take the derivative of the position function to get the velocity function: $\displaystyle v(t) = e^{3t$

Then take the derivative of the velocity function to get the acceleration function: $\displaystyle a(t) = 3e^{3t}$

Then, plug $\displaystyle t=0$ into the acceleration function: $\displaystyle a(0) = 3e^{3({\color{Blue} 0})}$

Therefore, the answer is: $\displaystyle 3$

In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function. Then find the derivative of the velocity function to get the acceleration function: $\displaystyle p''(t)= v'(t) = a(t)$

In this case, the position function is: $\displaystyle p(t) = (5t^3 - 8t^6)^2$

Then take the derivative of the position function to get the velocity function: $\displaystyle v(t) = 2(5t^3 - 8t^6)(15t^2 - 48t^5)$

Then take the derivative of the velocity function to get the acceleration function: $\displaystyle a(t) = 2(15t^2-48t^5) + 2(30t - 240t^4)(5t^3-8t^6)$

Then, plug $\displaystyle t=1$ into the acceleration function: $\displaystyle a(1) = 2(15(1)^2-48(1)^5) + 2(30(1) - 240(1)^4)(5(1)^3-8(1)^6)$

Therefore, the answer is: $\displaystyle 1194$

In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function. Then find the derivative of the velocity function to get the acceleration function: $\displaystyle p''(t)= v'(t) = a(t)$

In this case, the position function is: $\displaystyle p(t) = -\sin(2t)$

Then take the derivative of the position function to get the velocity function: $\displaystyle v(t) = -2\cos(2t)$

Then take the derivative of the velocity function to get the acceleration function: $\displaystyle a(t) = 4\sin(2t)$

Then, plug $\displaystyle t=1$ into the acceleration function: $\displaystyle a(1) = 4\sin(2({\color{Blue} 1}))$

Therefore, the answer is: $\displaystyle 3.64$

In order to find the acceleration of a certain point, you first find the derivative of the position function to get the velocity function. Then find the derivative of the velocity function to get the acceleration function: $\displaystyle p''(t)= v'(t) = a(t)$

In this case, the position function is: $\displaystyle p(t) = ln(7t^2 - \sin (2))$

Then take the derivative of the position function to get the velocity function: $\displaystyle v(t) = \frac{14t}{7t^2 - \sin 2}$

Then take the derivative of the velocity function to get the acceleration function: $\displaystyle a(t) = \frac{14(7t^2+\sin(2))}{(7t^2 - \sin 2)^2}$

Then, plug $\displaystyle t=1$ into the acceleration function: $\displaystyle a(1) = \frac{14(7({\color{Blue} 1})^2+\sin(2))}{(7({\color{Blue} 1})^2 - \sin 2)^2}$

Therefore, the answer is: $\displaystyle 2.98$

If v(t) is a function which models the velocity of a wave as a function of time, find the function which models the wave's acceleration.

$\displaystyle v(t)=5cos(t)+4t^2-5t$

We are given velocity and asked to find acceleration. To do so, we need to find the derivative of our velocity function.

$\displaystyle v'(t)=a(t)$

Recall that the derivative of cosine is negative sine, and that the derivative of any variable can be found by multiplying the term by the exponent and reducing the exponent by one.

$\displaystyle v'(t)=-5sin(t)+8t-5$

So we have:

$\displaystyle a(t)=-5sin(t)+8t-5$

If v(t) is a function which models the velocity of a wave as a function of time, find the the wave's acceleration when t=0.

$\displaystyle v(t)=5cos(t)+4t^2-5t$

We are given velocity and asked to find acceleration. To do so, we need to find the derivative of our velocity function.

$\displaystyle v'(t)=a(t)$

Recall that the derivative of cosine is negative sine, and that the derivative of any variable can be found by multiplying the term by the exponent and reducing the exponent by one.

$\displaystyle v'(t)=-5sin(t)+8t-5$

So we have:

$\displaystyle a(t)=-5sin(t)+8t-5$

We are not quite done yet however. We need to find a(0)

$\displaystyle a(0)=-5sin(0)+8(0)-5$

$\displaystyle a(0)=0+0-5=-5$

So our answer is -5

The instantaneous acceleration is represented by the second derivative of the positional equation. Let's first calculate the velocity then the acceleration. Begin by rewriting s(t) to make the differentiation easier:

s(t) = –44t² + 70√t = –44t² + 70(t)^1/2

v(t) = s'(t) = –88t + 70 * (1/2) * t^–1/2 = –88t + 35t^–1/2

a(t) = v'(t) = s''(t) = –88 - 70t^–3/2 = –88 -70/(t√t)

a(5) = –88 -70/(25√25) = -88 - 70/(5 * 5) = –88 - 70/25 = –88.82

To find acceleration we simply take the first derivative of velocity with respect to time:

$\displaystyle p(t)=3e^{4t}cos(2t)$

Using the product rule:

$\displaystyle \frac{\mathrm{d} p}{\mathrm{d} t}=12e^{4t}cos(2t)-6e^{4t}sin(2t)$

Factoring out a $\displaystyle 6e^{4t}$. 

$\displaystyle \frac{\mathrm{d} p}{\mathrm{d} t}=6e^{4t}(2cos(2t)-sin(2t))$