A body of mass `2kg` travels according to the law `x (t) = pt + qt^(2) + rt^(3)` where `p = 3ms^(-1),q = 4 ms^(-2)` and `r = 5ms^(-3)` . Find the force acting on the body at t=2 sec.

Options

A. 136N
B. 134N
C. 158N
D. 98N

Explanation

To find the force acting on the body at \( t = 2 \) seconds, we will follow these steps:

### Step 1: Write down the position function
The position of the body is given by the equation:
\[ x(t) = pt + qt^2 + rt^3 \]
where \( p = 3 \, \text{m/s} \), \( q = 4 \, \text{m/s}^2 \), and \( r = 5 \, \text{m/s}^3 \).

### Step 2: Differentiate to find velocity
To find the velocity \( v(t) \), we differentiate the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = p + 2qt + 3rt^2 \]
Substituting the values of \( p \), \( q \), and \( r \):
\[ v(t) = 3 + 2(4)t + 3(5)t^2 = 3 + 8t + 15t^2 \]

### Step 3: Differentiate to find acceleration
Next, we differentiate the velocity function to find the acceleration \( a(t) \):
\[ a(t) = \frac{dv}{dt} = 8 + 30t \]

### Step 4: Calculate acceleration at \( t = 2 \) seconds
Now, we substitute \( t = 2 \) seconds into the acceleration equation:
\[ a(2) = 8 + 30(2) = 8 + 60 = 68 \, \text{m/s}^2 \]

### Step 5: Calculate the force
Using Newton's second law, the force \( F \) acting on the body is given by:
\[ F = m \cdot a \]
where \( m = 2 \, \text{kg} \) (mass of the body) and \( a = 68 \, \text{m/s}^2 \) (acceleration at \( t = 2 \) seconds):
\[ F = 2 \cdot 68 = 136 \, \text{N} \]

### Final Answer
The force acting on the body at \( t = 2 \) seconds is \( 136 \, \text{N} \).

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