The velocity of a diver just before hitting the water is -8.8 m/s, where the minus sign indicates that her motion is directly downward. What is her displacement during the last 0.93 s of the dive?

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To find the displacement of the diver during the last 0.93 seconds of the dive, we can use the equation:

Displacement = initial velocity × time + (1/2) × acceleration × time²

First, let's identify the given values:
Initial velocity (u) = -8.8 m/s (downward)
Time (t) = 0.93 s
Acceleration (a) = ?

Since the diver's motion is directly downward, we can assume that the acceleration due to gravity (g) is acting on the diver. The value of gravity on Earth is approximately -9.8 m/s² (pointing downwards).

Substituting the values into the equation, we have:
Displacement = -8.8 m/s × 0.93 s + (1/2) × (-9.8 m/s²) × (0.93 s)²

Simplifying the equation, we get:
Displacement = -8.184 m + (-4.7175 m) = -12.9015 m (rounded to three decimal places)

Therefore, the displacement of the diver during the last 0.93 seconds of the dive is approximately -12.9015 meters. Since the displacement is negative, it indicates that the diver moved downwards.