Final answer:
The distance traveled by the person's hand when it slaps the leg and comes to rest from an initial speed of 4.15 m/s in 265 ms is approximately 0.549 m.
Step-by-step explanation:
To calculate the distance traveled by a person's hand when they slap their leg and bring it to rest from an initial speed of 4.15 m/s in 265 ms (0.265 seconds), we can use the equation for constant acceleration:
distance = initial speed x time + (1/2) x acceleration x time2.
First, we need to calculate the acceleration required to bring the hand to rest. This can be done using the formula: acceleration = (final speed - initial speed) / time, which in this case is (0 m/s - 4.15 m/s) / 0.265 s. The acceleration is therefore -15.66 m/s2 (the negative indicates deceleration).
Inserting these values into the distance calculation, we get:
distance = 4.15 m/s x 0.265 s + (1/2) x (-15.66 m/s2) x (0.265 s)2
After carrying out the calculations, the distance traveled is approximately 0.549 m (rounded to three decimal places).