A body of mass 10 kg is upon by a force given by equation `F=(3t^(2)-30)` newtons. The initial velocity of the body is 10 m/s. The velocity of the body after 5 sec. is

Options

A. 4.5 m/s
B. 6 m/s
C. 7.5 m/s
D. 5 m/s

Answer

To solve the problem step by step, we will follow the given information and apply the relevant physics concepts.

### Step 1: Identify the given information
- Mass of the body, \( m = 10 \, \text{kg} \)
- Force acting on the body, \( F(t) = 3t^2 - 30 \, \text{N} \)
- Initial velocity, \( v_0 = 10 \, \text{m/s} \)
- Time duration, \( t = 5 \, \text{s} \)

### Step 2: Relate force to acceleration
Using Newton's second law, we know that:
\[ F = m \cdot a \]
where \( a \) is the acceleration. Rearranging gives:
\[ a = \frac{F}{m} \]
Substituting the expression for force:
\[ a(t) = \frac{3t^2 - 30}{10} \]

### Step 3: Express acceleration in terms of velocity
Since acceleration is the rate of change of velocity, we can express it as:
\[ a = \frac{dv}{dt} \]
Thus, we have:
\[ \frac{dv}{dt} = \frac{3t^2 - 30}{10} \]

### Step 4: Separate variables and integrate
We can separate the variables:
\[ dv = \frac{3t^2 - 30}{10} dt \]
Now, we will integrate both sides. The limits for \( v \) will be from \( v_0 = 10 \) to \( v \), and for \( t \) from \( 0 \) to \( 5 \):
\[ \int_{10}^{v} dv = \int_{0}^{5} \frac{3t^2 - 30}{10} dt \]

### Step 5: Calculate the right-hand side integral
Calculating the integral on the right:
\[ \int (3t^2 - 30) dt = t^3 - 30t \]
Evaluating from \( 0 \) to \( 5 \):
\[ \left[ \frac{1}{10}(t^3 - 30t) \right]_{0}^{5} = \frac{1}{10} \left[ (5^3 - 30 \cdot 5) - (0 - 0) \right] \]
Calculating:
\[ = \frac{1}{10} \left[ 125 - 150 \right] = \frac{1}{10} \left[ -25 \right] = -2.5 \]

### Step 6: Solve for the final velocity
Now substituting back into the equation:
\[ v - 10 = -2.5 \]
Thus,
\[ v = 10 - 2.5 = 7.5 \, \text{m/s} \]

### Final Answer
The velocity of the body after 5 seconds is \( 7.5 \, \text{m/s} \).

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