An onject intially att rest moves along x-axis subjected to an acceleration which veries with time according to relation `a=2t+5`. Its velocity after 2 seconds will be :-

Options

A. `18 ms^(-1)`
B. `9 ms^(-1)`
C. `12 ms^(-1)`
D. `14 ms^(-1)`

Answer

To find the velocity of the object after 2 seconds, we can follow these steps:

### Step 1: Understand the given information
The acceleration \( a \) of the object is given by the equation:
\[ a = 2t + 5 \]
The object starts from rest, which means its initial velocity \( v_0 = 0 \).

### Step 2: Relate acceleration to velocity
Acceleration is the rate of change of velocity with respect to time. Therefore, we can express this relationship as:
\[ a = \frac{dv}{dt} \]
We can substitute the expression for acceleration into this equation:
\[ \frac{dv}{dt} = 2t + 5 \]

### Step 3: Integrate the acceleration to find velocity
To find the velocity \( v \), we need to integrate the acceleration with respect to time:
\[ v = \int (2t + 5) \, dt \]
This can be separated into two integrals:
\[ v = \int 2t \, dt + \int 5 \, dt \]

### Step 4: Perform the integration
Now we can perform the integration:
1. The integral of \( 2t \) is \( t^2 \).
2. The integral of \( 5 \) is \( 5t \).

Thus, we have:
\[ v = t^2 + 5t + C \]
where \( C \) is the constant of integration.

### Step 5: Apply the initial condition
Since the object starts from rest, we know that when \( t = 0 \), \( v = 0 \):
\[ 0 = 0^2 + 5(0) + C \implies C = 0 \]
So the expression for velocity simplifies to:
\[ v = t^2 + 5t \]

### Step 6: Calculate the velocity at \( t = 2 \) seconds
Now we can substitute \( t = 2 \) seconds into the velocity equation:
\[ v = (2)^2 + 5(2) \]
Calculating this gives:
\[ v = 4 + 10 = 14 \, \text{m/s} \]

### Final Answer
The velocity of the object after 2 seconds is:
\[ \boxed{14 \, \text{m/s}} \]