Question about acceleration equation that is derived from place equation.

Question

• $a=4$

• ${\bf R}(t)=7\sin(at)\hat{{\bf x}}+4e^{-8t}\hat{{\bf y}}+8t^{3}\hat{{\bf z}}$

how do I find the acceleration at time $t = 0.27778$

I know that the third derivative is: $ \vec {R^{(3)}(t)} = -7a^3\cos(at)\hat x -2048e^{-8t}\hat y +48\hat z$

Now, how do I take it from here? and how do I calculate it?

Answer 1

If $R(t)$ is the position vector at time $t$, then $R'(t)$ is the velocity vector at time $t$ and $R''(t)$ is the acceleration vector at time $t$. Here the derivatives are taken component-wise.

The third derivative $R^{(3)}(t)$ is sometimes called jerk, as I recall.

Answer 2

The acceleration is the second derivative of the position vector, not the third.

Just find $\overrightarrow {{R^{(2)}}(t)}$ then substitute $t=0.27778$.