After a jump, a skydiver reaches terminal speed in 10 seconds. Does he gain more speed during the first second of fall or the ninth second of fall? which of these two seconds (1st or 9th) will he fall the greater distance?
According to the drag equation,
a = g - k/m v^2
and the diver approaches terminal velocity asymptotically. Sounds to me like he gains speed more slowly the longer he falls.
A little web searching should clear it up.
The skydiver gains more speed during the first second of the fall. This is because the acceleration due to gravity, which causes the skydiver to gain speed, remains constant throughout the entire fall until terminal velocity is reached.
During the first second, the skydiver starts from rest and accelerates due to gravity. However, in the ninth second, the skydiver has already reached terminal velocity and is no longer accelerating significantly.
In terms of falling distance, the greater distance will be covered during the ninth second of the fall. This is because the skydiver spends more time at terminal velocity, covering a greater distance compared to the first second of rapid acceleration.
To determine whether the skydiver gains more speed during the first or the ninth second of the fall, we need to understand the concept of terminal speed. Terminal speed is the maximum velocity an object can achieve while falling due to the balance between the force of gravity pulling the object downward and the air resistance pushing against it.
When the skydiver first jumps out of the plane, he starts accelerating due to the force of gravity since there is no significant air resistance acting on him. As he falls and gains speed, the air resistance gradually increases, opposing the force of gravity.
At some point, the force of air resistance will become equal to the force of gravity, causing the skydiver to reach terminal speed. At terminal speed, the net force experienced by the skydiver becomes zero, which means that his acceleration stops, and his speed remains constant.
Based on this understanding, during the first second of the fall, the skydiver is still accelerating due to the force of gravity. Therefore, he gains more speed during the first second as compared to the ninth second when he has already reached terminal speed.
Now, let's consider the distance traveled by the skydiver during the first and ninth seconds. Since the skydiver is free-falling during these seconds, we can use the equations of motion under constant acceleration to calculate the distance traveled.
During the first second of the fall, the skydiver is accelerating, so the distance traveled can be determined using the equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
Since the skydiver starts from rest, the initial velocity is zero. The acceleration due to gravity is approximately 9.8 m/s^2. Plugging in the values, we get:
distance_1st_second = 0 * 1 + (1/2) * 9.8 * 1^2
= 0 + (1/2) * 9.8 * 1
= 4.9 meters
Therefore, during the first second of the fall, the skydiver will travel a distance of 4.9 meters.
On the other hand, during the ninth second of the fall, the skydiver has already reached terminal speed, meaning his velocity remains constant. In this case, the distance traveled can be determined using the equation:
distance = velocity * time
Since the velocity is constant during the ninth second, we can simply multiply the terminal speed by the time:
distance_9th_second = terminal speed * 1
= terminal speed
Therefore, during the ninth second of the fall, the skydiver will travel a distance equal to his terminal speed.
In conclusion, the skydiver gains more speed during the first second of the fall, but the distance traveled during the ninth second will be greater if the skydiver continues to fall for that duration.