If the relation between acceleration and time for an object is given by
`a=2t+4t^(2)`
Calculate the position of object from the origin at t = 4 s. Assume the object to be at rest at `t=0`.

Explanation

To solve the problem step by step, we will follow the process of integrating the acceleration to find the velocity, and then integrating the velocity to find the position.

### Step 1: Write the given equation for acceleration
The acceleration \( a \) is given by:
\[ a = 2t + 4t^2 \]

### Step 2: Relate acceleration to velocity
Acceleration is the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Thus, we can write:
\[ \frac{dv}{dt} = 2t + 4t^2 \]

### Step 3: Integrate to find velocity
We will integrate both sides with respect to \( t \):
\[ dv = (2t + 4t^2) dt \]
Integrating gives:
\[ v = \int (2t + 4t^2) dt \]
Calculating the integral:
\[ v = \int 2t \, dt + \int 4t^2 \, dt = t^2 + \frac{4}{3}t^3 + C \]
Since the object is at rest at \( t = 0 \), we have \( v(0) = 0 \):
\[ 0 = 0 + 0 + C \implies C = 0 \]
Thus, the velocity equation simplifies to:
\[ v = t^2 + \frac{4}{3}t^3 \]

### Step 4: Relate velocity to position
Velocity is also the rate of change of position:
\[ v = \frac{dx}{dt} \]
Thus, we can write:
\[ \frac{dx}{dt} = t^2 + \frac{4}{3}t^3 \]

### Step 5: Integrate to find position
We will integrate both sides with respect to \( t \):
\[ dx = \left(t^2 + \frac{4}{3}t^3\right) dt \]
Integrating gives:
\[ x = \int \left(t^2 + \frac{4}{3}t^3\right) dt \]
Calculating the integral:
\[ x = \int t^2 \, dt + \int \frac{4}{3}t^3 \, dt = \frac{t^3}{3} + \frac{4}{12}t^4 + C \]
Simplifying gives:
\[ x = \frac{t^3}{3} + \frac{1}{3}t^4 + C \]
Since the object starts from the origin at \( t = 0 \), we have \( x(0) = 0 \):
\[ 0 = 0 + 0 + C \implies C = 0 \]
Thus, the position equation simplifies to:
\[ x = \frac{t^3}{3} + \frac{1}{3}t^4 \]

### Step 6: Calculate position at \( t = 4 \) seconds
Now we substitute \( t = 4 \) seconds into the position equation:
\[ x = \frac{4^3}{3} + \frac{1}{3}(4^4) \]
Calculating each term:
\[ x = \frac{64}{3} + \frac{256}{3} = \frac{320}{3} \approx 106.67 \text{ meters} \]

### Final Answer
The position of the object from the origin at \( t = 4 \) seconds is approximately:
\[ x \approx 106.67 \text{ meters} \]