For any particle moving with some velocity `(vecv)` & acceleration `(veca)`, tangential acceleration & normal acceleration are defined as follows.
Tangential acceleration - The component of acceleration in the direction of velocity.
Normal acceleration - The component of acceleration in the direction perpendicular to velocity.
If at a given instant, velocity & acceleration of a particle are given by .
`vecc=4hati +3hatj`
`veca=10hati+15hatj+20hatk`
Find the tangential acceleration of the particle at the given instant :-

Options

A. `17(4hati+3hatj)`
B. `(17)/(5)(4hati+3hatj)`
C. `17(4hati-3hatj)`
D. `(17)/(5)(4hati-3hatj)`

Answer

To find the tangential acceleration of the particle given its velocity and acceleration vectors, we can follow these steps:

### Step 1: Identify the given vectors
We have the following vectors:
- Velocity vector:
\[ \vec{v} = 4 \hat{i} + 3 \hat{j} \]
- Acceleration vector:
\[ \vec{a} = 10 \hat{i} + 15 \hat{j} + 20 \hat{k} \]

### Step 2: Calculate the magnitude of the velocity vector
The magnitude of the velocity vector \(|\vec{v}|\) can be calculated using the formula:
\[ |\vec{v}| = \sqrt{(v_x)^2 + (v_y)^2} \]
where \(v_x\) and \(v_y\) are the components of the velocity vector.

Substituting the values:
\[ |\vec{v}| = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 3: Calculate the dot product of acceleration and velocity vectors
The tangential acceleration \(a_t\) can be found using the formula:
\[ a_t = \frac{\vec{a} \cdot \vec{v}}{|\vec{v}|} \]
First, we need to calculate the dot product \(\vec{a} \cdot \vec{v}\):
\[ \vec{a} \cdot \vec{v} = (10 \hat{i} + 15 \hat{j} + 20 \hat{k}) \cdot (4 \hat{i} + 3 \hat{j}) \]
Calculating the dot product:
\[ \vec{a} \cdot \vec{v} = (10 \cdot 4) + (15 \cdot 3) + (20 \cdot 0) = 40 + 45 + 0 = 85 \]

### Step 4: Calculate the tangential acceleration
Now, substituting the values into the tangential acceleration formula:
\[ a_t = \frac{85}{5} = 17 \]

### Step 5: Write the final result
The tangential acceleration of the particle at the given instant is:
\[ \boxed{17} \]