The displacement y (in metres) of a body varies with time t ( in seconds ) as `y= (-2)/(3) t^(2)+16t-12` . How long does the body take to come to rest ?

Options

A. 16 sec
B. 12 sec
C. 10 sec
D. 8sec

Answer

To find out how long the body takes to come to rest, we need to determine when the velocity of the body becomes zero. The displacement \( y \) of the body is given by the equation:

\[ y = -\frac{2}{3} t^2 + 16t - 12 \]

### Step 1: Find the velocity function
The velocity \( v \) is the derivative of the displacement \( y \) with respect to time \( t \). We can find the velocity by differentiating the displacement function:

\[ v = \frac{dy}{dt} = \frac{d}{dt} \left( -\frac{2}{3} t^2 + 16t - 12 \right) \]

### Step 2: Differentiate the displacement
Using the power rule of differentiation, we differentiate each term:

1. The derivative of \( -\frac{2}{3} t^2 \) is \( -\frac{2}{3} \cdot 2t = -\frac{4}{3} t \).
2. The derivative of \( 16t \) is \( 16 \).
3. The derivative of \( -12 \) is \( 0 \).

Combining these, we have:

\[ v = -\frac{4}{3} t + 16 \]

### Step 3: Set the velocity to zero
To find when the body comes to rest, we set the velocity \( v \) to zero:

\[ -\frac{4}{3} t + 16 = 0 \]

### Step 4: Solve for \( t \)
Now, we solve for \( t \):

1. Add \( \frac{4}{3} t \) to both sides:

\[ 16 = \frac{4}{3} t \]

2. Multiply both sides by \( \frac{3}{4} \):

\[ t = 16 \cdot \frac{3}{4} = 12 \]

### Conclusion
The body takes \( t = 12 \) seconds to come to rest.

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