The velocity (in m/s) of an object changes from ` vecv_(1) = 10 hati + 4hatj + 2hatk " to " vecv_(2) = 4hati + 2hatj + 3hatk` in 5 second. Find the magnitude of average acceleration.

Answer

To find the magnitude of average acceleration when the velocity of an object changes from \( \vec{v}_1 = 10 \hat{i} + 4 \hat{j} + 2 \hat{k} \) to \( \vec{v}_2 = 4 \hat{i} + 2 \hat{j} + 3 \hat{k} \) in 5 seconds, we can follow these steps:

### Step 1: Calculate the change in velocity
The change in velocity \( \Delta \vec{v} \) is given by:
\[ \Delta \vec{v} = \vec{v}_2 - \vec{v}_1 \]
Substituting the given values:
\[ \Delta \vec{v} = (4 \hat{i} + 2 \hat{j} + 3 \hat{k}) - (10 \hat{i} + 4 \hat{j} + 2 \hat{k}) \]
Calculating this gives:
\[ \Delta \vec{v} = (4 - 10) \hat{i} + (2 - 4) \hat{j} + (3 - 2) \hat{k} = -6 \hat{i} - 2 \hat{j} + 1 \hat{k} \]

### Step 2: Calculate the average acceleration
The average acceleration \( \vec{a} \) is defined as the change in velocity divided by the time interval \( T \):
\[ \vec{a} = \frac{\Delta \vec{v}}{T} \]
Substituting the values:
\[ \vec{a} = \frac{-6 \hat{i} - 2 \hat{j} + 1 \hat{k}}{5} \]
Calculating this gives:
\[ \vec{a} = -\frac{6}{5} \hat{i} - \frac{2}{5} \hat{j} + \frac{1}{5} \hat{k} \]

### Step 3: Calculate the magnitude of average acceleration
The magnitude of the average acceleration \( |\vec{a}| \) is given by:
\[ |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \]
Where \( a_x = -\frac{6}{5} \), \( a_y = -\frac{2}{5} \), and \( a_z = \frac{1}{5} \). Thus:
\[ |\vec{a}| = \sqrt{\left(-\frac{6}{5}\right)^2 + \left(-\frac{2}{5}\right)^2 + \left(\frac{1}{5}\right)^2} \]
Calculating each term:
\[ |\vec{a}| = \sqrt{\frac{36}{25} + \frac{4}{25} + \frac{1}{25}} = \sqrt{\frac{41}{25}} = \frac{\sqrt{41}}{5} \]
Calculating \( \sqrt{41} \approx 6.4 \):
\[ |\vec{a}| \approx \frac{6.4}{5} \approx 1.28 \, \text{m/s}^2 \]

### Final Answer:
The magnitude of average acceleration is approximately \( 1.28 \, \text{m/s}^2 \).

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