a skydiver steps from a helicopter and falls for 5 seconds before reaching her terminal velocity.during this 5 second interval, her acceleration;
To find the acceleration of the skydiver during the 5-second interval, we can use the equation of motion:
v = u + at
Where:
v is the final velocity (terminal velocity)
u is the initial velocity (0 m/s as the skydiver starts from rest)
a is the acceleration
t is the time taken (5 seconds)
In this case, since the skydiver reaches her terminal velocity, her final velocity is constant. Hence, we have:
v = 0 + at
Rearranging the equation, we get:
a = v/t
Since the equation is asking for the acceleration specifically during the 5-second interval, we cannot determine the value without additional information. The value of the terminal velocity needs to be known in order to calculate the acceleration.
To determine the skydiver's acceleration during the 5-second interval, we need to consider her motion and use the equations of motion.
The skydiver is falling under the influence of gravity, so her motion can be described by the equation:
y = y0 + v0t + (1/2)at^2
where:
y is the position or displacement of the skydiver (height)
y0 is the initial position (height) of the skydiver (in this case, from the helicopter)
v0 is the initial velocity of the skydiver (which we assume to be zero since she stepped off the helicopter)
t is the time elapsed
a is the acceleration
Since the skydiver is falling vertically downward, we can consider the positive direction as downward. Thus, the equation can be simplified to:
y = y0 - (1/2)gt^2
where g is the acceleration due to gravity, approximately 9.8 m/s^2.
Since the skydiver reaches her terminal velocity, it means her acceleration becomes zero after this point. Terminal velocity occurs when the force of air resistance becomes equal to the force of gravity, resulting in no further acceleration.
Therefore, during the 5-second interval, we can say her acceleration is constant and equal to the acceleration due to gravity (g).
Hence, the skydiver's acceleration during the 5-second interval is approximately 9.8 m/s^2.
v =g•t = 9.8•5 = 49 m/s.
a = g