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

Options

A. 136 N
B. 134 N
C. 158 N
D. 68 N

Explanation

To find the force acting on a body of mass 2 kg that travels according to the law \( x(t) = pt + qt^2 + rt^3 \) where \( p = 3 \, \text{ms}^{-1} \), \( q = 4 \, \text{ms}^{-2} \), and \( r = 5 \, \text{ms}^{-3} \), we will follow these steps:

### Step 1: Identify the given values
- Mass \( m = 2 \, \text{kg} \)
- Time \( t = 2 \, \text{s} \)
- Coefficients: \( p = 3 \, \text{ms}^{-1} \), \( q = 4 \, \text{ms}^{-2} \), \( r = 5 \, \text{ms}^{-3} \)

### Step 2: Write the position function
The position function is given as:
\[ x(t) = pt + qt^2 + rt^3 \]
Substituting the values of \( p \), \( q \), and \( r \):
\[ x(t) = 3t + 4t^2 + 5t^3 \]

### Step 3: Find the velocity function
The velocity \( v(t) \) is the first derivative of the position function \( x(t) \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t + 4t^2 + 5t^3) \]
Calculating the derivative:
\[ v(t) = 3 + 8t + 15t^2 \]

### Step 4: Find the acceleration function
The acceleration \( a(t) \) is the derivative of the velocity function \( v(t) \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(3 + 8t + 15t^2) \]
Calculating the derivative:
\[ a(t) = 0 + 8 + 30t = 8 + 30t \]

### Step 5: Calculate acceleration at \( t = 2 \, \text{s} \)
Substituting \( t = 2 \) into the acceleration function:
\[ a(2) = 8 + 30(2) = 8 + 60 = 68 \, \text{ms}^{-2} \]

### Step 6: Calculate the force
Using Newton's second law, the force \( F \) is given by:
\[ F = m \cdot a \]
Substituting the mass and the acceleration:
\[ F = 2 \, \text{kg} \cdot 68 \, \text{ms}^{-2} = 136 \, \text{N} \]

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

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