The velocity of a diver just before hitting the water is `-10.1 m//s`, where the minus sign indicates that her motion is directly downward. What is her displacement during the last 1.20 s of the dive ?

Options

A. `-5.06m`
B. `-7.06m`
C. `-12.1`m
D. `-4.27` m

Answer

To find the displacement of the diver during the last 1.20 seconds of her dive, we can use the equations of motion under constant acceleration. Here’s the step-by-step solution:

### Step 1: Identify known values
- Final velocity (V) just before hitting the water: \( V = -10.1 \, \text{m/s} \) (downward)
- Time interval (t): \( t = 1.20 \, \text{s} \)
- Acceleration (a): \( a = -9.8 \, \text{m/s}^2 \) (due to gravity, downward)

### Step 2: Use the equation of motion to find initial velocity (u)
We can use the equation:
\[ V = u + at \]
Rearranging gives us:
\[ u = V - at \]
Substituting the known values:
\[ u = -10.1 \, \text{m/s} - (-9.8 \, \text{m/s}^2 \times 1.20 \, \text{s}) \]
Calculating:
\[ u = -10.1 \, \text{m/s} + 11.76 \, \text{m/s} = 1.66 \, \text{m/s} \]

### Step 3: Use the equation of motion to find displacement (s)
Now we can use the equation:
\[ s = \frac{V^2 - u^2}{2a} \]
Substituting the values:
\[ s = \frac{(-10.1)^2 - (1.66)^2}{2 \times (-9.8)} \]
Calculating the squares:
\[ s = \frac{102.01 - 2.7556}{-19.6} \]
\[ s = \frac{99.2544}{-19.6} \approx -5.06 \, \text{m} \]

### Conclusion
The displacement of the diver during the last 1.20 seconds of the dive is approximately:
\[ \text{Displacement} = -5.06 \, \text{m} \]

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