A skydiver is falling straight down at 60 m/s when he opens his parachute and slows to 8.3 m/s in 4.0 s. What is the average acceleration of the skydiver during those 4.0 s? (Assume that "up" is the positive direction.)
Average acceleration
= (velocity change)/(time)
You are given the final and initial velocity. The V change is
8.3 - 60 = -51.7 m/s
To find the average acceleration of the skydiver, we can use the formula:
average acceleration = change in velocity / time interval
In this case, the change in velocity is the difference between the final velocity and the initial velocity, while the time interval is given as 4.0 s.
The initial velocity is 60 m/s (when the skydiver is falling straight down), and the final velocity is 8.3 m/s (after the parachute opened). Let's now substitute these values into the formula:
average acceleration = (8.3 m/s - 60 m/s) / 4.0 s
To simplify the calculation, we need to express the final velocity with the proper sign convention. Since "up" is considered the positive direction, the final velocity will be positive. Therefore, the expression becomes:
average acceleration = (8.3 m/s - (-60 m/s)) / 4.0 s
Now, we can simplify the numerator:
average acceleration = (8.3 m/s + 60 m/s) / 4.0 s
Adding the two velocities:
average acceleration = (68.3 m/s) / 4.0 s
Dividing the numerator by the denominator:
average acceleration ≈ 17.1 m/s²
Therefore, the average acceleration of the skydiver during those 4.0 s is approximately 17.1 m/s².