A particle moves along the space curve `vec(r)=(t^(2)+t)hat(i)+(3t-2)hat(j)+(2t^(3)-4t^(2))hat(k)`. (t in sec, r in m ) Find at time `t=2` the (a) velocity, (b) acceleration, (c) speed or magnitude of velocity and (d) magnitude of acceleration.

Explanation

To solve the problem step by step, we need to find the velocity, acceleration, speed (magnitude of velocity), and magnitude of acceleration of a particle moving along the given space curve at time \( t = 2 \) seconds.

The position vector of the particle is given by:
\[ \vec{r}(t) = (t^2 + t) \hat{i} + (3t - 2) \hat{j} + (2t^3 - 4t^2) \hat{k} \]

### Step 1: Find the Velocity
The velocity vector \( \vec{v}(t) \) is the derivative of the position vector \( \vec{r}(t) \) with respect to time \( t \).

\[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}[(t^2 + t) \hat{i} + (3t - 2) \hat{j} + (2t^3 - 4t^2) \hat{k}] \]

Calculating the derivatives:
- For \( \hat{i} \): \( \frac{d}{dt}(t^2 + t) = 2t + 1 \)
- For \( \hat{j} \): \( \frac{d}{dt}(3t - 2) = 3 \)
- For \( \hat{k} \): \( \frac{d}{dt}(2t^3 - 4t^2) = 6t^2 - 8t \)

Thus, the velocity vector is:
\[ \vec{v}(t) = (2t + 1) \hat{i} + 3 \hat{j} + (6t^2 - 8t) \hat{k} \]

Now, substituting \( t = 2 \):
\[ \vec{v}(2) = (2(2) + 1) \hat{i} + 3 \hat{j} + (6(2^2) - 8(2)) \hat{k} \]
\[ = (4 + 1) \hat{i} + 3 \hat{j} + (24 - 16) \hat{k} \]
\[ = 5 \hat{i} + 3 \hat{j} + 8 \hat{k} \]

### Step 2: Find the Acceleration
The acceleration vector \( \vec{a}(t) \) is the derivative of the velocity vector \( \vec{v}(t) \).

\[ \vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}[(2t + 1) \hat{i} + 3 \hat{j} + (6t^2 - 8t) \hat{k}] \]

Calculating the derivatives:
- For \( \hat{i} \): \( \frac{d}{dt}(2t + 1) = 2 \)
- For \( \hat{j} \): \( \frac{d}{dt}(3) = 0 \)
- For \( \hat{k} \): \( \frac{d}{dt}(6t^2 - 8t) = 12t - 8 \)

Thus, the acceleration vector is:
\[ \vec{a}(t) = 2 \hat{i} + 0 \hat{j} + (12t - 8) \hat{k} \]

Now, substituting \( t = 2 \):
\[ \vec{a}(2) = 2 \hat{i} + 0 \hat{j} + (12(2) - 8) \hat{k} \]
\[ = 2 \hat{i} + 0 \hat{j} + (24 - 8) \hat{k} \]
\[ = 2 \hat{i} + 0 \hat{j} + 16 \hat{k} \]

### Step 3: Find the Speed (Magnitude of Velocity)
The speed is the magnitude of the velocity vector \( \vec{v}(2) \):
\[ |\vec{v}(2)| = \sqrt{(5)^2 + (3)^2 + (8)^2} \]
\[ = \sqrt{25 + 9 + 64} = \sqrt{98} = 7\sqrt{2} \text{ m/s} \]

### Step 4: Find the Magnitude of Acceleration
The magnitude of the acceleration vector \( \vec{a}(2) \):
\[ |\vec{a}(2)| = \sqrt{(2)^2 + (0)^2 + (16)^2} \]
\[ = \sqrt{4 + 0 + 256} = \sqrt{260} = 2\sqrt{65} \text{ m/s}^2 \]

### Summary of Results
- (a) Velocity at \( t = 2 \): \( \vec{v}(2) = 5 \hat{i} + 3 \hat{j} + 8 \hat{k} \) m/s
- (b) Acceleration at \( t = 2 \): \( \vec{a}(2) = 2 \hat{i} + 0 \hat{j} + 16 \hat{k} \) m/s²
- (c) Speed at \( t = 2 \): \( 7\sqrt{2} \) m/s
- (d) Magnitude of acceleration at \( t = 2 \): \( 2\sqrt{65} \) m/s²