The velocity of a diver just before hitting the water is -8.4 m/s, where the minus sign indicates that her motion is directly downward. What is her displacement during the last 0.82 s of the dive?
This problem with different numbers has been done on here before, but when i followed the steps i did not get a correct answer
To determine the displacement of the diver during the last 0.82 seconds of the dive, we can use the equation of motion:
displacement = initial velocity × time + (1/2) × acceleration × time^2
Given:
Initial velocity (u) = -8.4 m/s (downward)
Time (t) = 0.82 s
Acceleration (a) is not provided in the question, but we can assume that the acceleration is due to gravity, which is approximately constant at -9.8 m/s^2 (downward).
Since the motion is directly downward, we can take acceleration as negative.
Now, let's plug in the values into the equation:
displacement = -8.4 m/s × 0.82 s + (1/2)(-9.8 m/s^2)(0.82 s)^2
Calculating further:
displacement = -6.888 m + (1/2) × -9.8 m/s^2 × (0.6724 s^2)
displacement = -6.888 m - 3.308 m
displacement = -10.196 m
The displacement of the diver during the last 0.82 seconds of the dive is approximately -10.2 m (downward).
If you didn't get the correct answer when following the steps, please double-check the calculations and ensure that you have used the correct values for the initial velocity, time, and acceleration.