Question

Use the given acceleration vector to find the velocity and position vectors. Then find the position at time t=2: a(t) = −32k, v(0) = 3i − 2j + k, r(0) = 5j + 2k

Answer 1

Answer: the velocity and position vector is given by;

v(t) = 3i + 2j + (1-32t)k

r(t) = (3t)i + (5-2t)j + (2+t-16t^2)k

The position at time t=2; r(2)

r(2) = 6i + j - 60k

Step-by-step explanation:

Shown in the attachment.

Answer 2

Answer:

The velocity vector is $\vec{v(t)}=3 \hat{i}-2 \hat{j}+(-32t+1)\hat{k}$ and the position vector is $\vec{r(t)}=3t \hat{i}-(2 t-5)\hat{j}+(-32\frac{t^2 }{2} +t+2)\hat{k}}$

When t = 2  the position is $\vec{r(2)}=6 \hat{i}+\hat{j}-60\hat{k}}$

Step-by-step explanation:

To find the position function, you should integrate twice, each time using one of the initial conditions to solve for the constant of integration. The velocity vector is

${\vec{v(t)}}=\int\limits {\vec{a(t)}} \, dt=\int\limits {-32 \hat{k}} \, dt\\\\\vec{v(t)}=-32t \hat{k}+C$

where $C=C_1 \hat{i}+C_2 \hat{j}+C_3 \hat{k}$.

Letting t = 0 and applying the initial condition $\vec{v(0)} = 3\hat{i} - 2\hat{j} + \hat{k}$

$\vec{v(0)}=-32(0) \hat{k}+C\\\vec{v(0)}=C\\\\3\hat{i} - 2\hat{j} + \hat{k}=C_1 \hat{i}+C_2 \hat{j}+C_3 \hat{k}$

Therefore,

$C_1=3, C_2=-2,C_3=1$

So, the velocity at any time t is

$\vec{v(t)}=-32t \hat{k}+3 \hat{i}-2 \hat{j}+1\hat{k}\\\vec{v(t)}=3 \hat{i}-2 \hat{j}+(-32t+1)\hat{k}$

Integrating once more produces

$\vec{r(t)}=\int\limits {\vec{v(t)}} \, dt=\int\limits {3 \hat{i}-2 \hat{j}+(-32t+1)\hat{k}} \, dt\\\\\vec{r(t)}=3t \hat{i}-2 t\hat{j}+(-32\frac{t^2 }{2} +t)\hat{k}}+C$

Letting t = 0 and applying the initial condition $\vec{r(0)} = 5\hat{j} + 2\hat{k}$

$\vec{r(0)}=3(0) \hat{i}-2 (0)\hat{j}+(-32\frac{(0)^2 }{2} +(0))\hat{k}}+C\\\vec{r(0)}=C\\\\5\hat{j} + 2\hat{k}=C_1 \hat{i}+C_2 \hat{j}+C_3 \hat{k}$

Therefore,

$C_1=0, C_2=5,C_3=2$

The position vector is

$\vec{r(t)}=3t \hat{i}-2 t\hat{j}+(-32\frac{t^2 }{2} +t)\hat{k}}+5 \hat{j}+2 \hat{k}\\\\\vec{r(t)}=3t \hat{i}-(2 t-5)\hat{j}+(-32\frac{t^2 }{2} +t+2)\hat{k}}$

When t = 2  the position is

$\vec{r(2)}=3(2) \hat{i}-(2(2)-5)\hat{j}+(-32\frac{(2)^2 }{2} +(2)+2)\hat{k}}\\\vec{r(2)}=6 \hat{i}+\hat{j}-60\hat{k}}$