The motion of a particle is described by the equation `x = a+bt^(2)` where `a = 15` cm and `b = 3 cm//s`. Its instantaneous velocity at time 3 sec will be

Options

A. `36 cm//sec`
B. `18 cm//sec`
C. `16 cm//sec`
D. `32 cm//sec`

Answer

To find the instantaneous velocity of a particle described by the equation \( x = a + bt^2 \), where \( a = 15 \) cm and \( b = 3 \) cm/s, we can follow these steps:

### Step 1: Understand the given equation
The position \( x \) of the particle as a function of time \( t \) is given by:
\[ x = a + bt^2 \]
where \( a = 15 \) cm and \( b = 3 \) cm/s.

### Step 2: Differentiate the position function
To find the instantaneous velocity \( v \), we need to differentiate the position function \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} \]
Differentiating \( x = a + bt^2 \):
\[ \frac{dx}{dt} = 0 + 2bt \]
Thus, the velocity \( v \) is:
\[ v = 2bt \]

### Step 3: Substitute the value of \( b \)
Now, substitute the value of \( b = 3 \) cm/s into the velocity equation:
\[ v = 2 \cdot 3 \cdot t = 6t \]

### Step 4: Calculate the velocity at \( t = 3 \) seconds
Now we need to find the instantaneous velocity at \( t = 3 \) seconds:
\[ v = 6 \cdot 3 = 18 \text{ cm/s} \]

### Conclusion
The instantaneous velocity of the particle at \( t = 3 \) seconds is:
\[ \boxed{18 \text{ cm/s}} \]

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