A sky diver jumps from a reasonable height above the ground. The air resistance she experiences is proportional to her velocity, and the constant of proportionality is 0.23. It can be shown that the downward velocity of the sky diver at time t is given by
v(t) = 180(1 − e−0.23t)
where t is measured in seconds and v(t) is measured in feet per second (ft/s).
(a) Find the initial velocity of the sky diver
(b) Find the velocity after 5 s and after 10 s. (Round your answers to one decimal place.)
after 5 s = ft/s
after 10 s = ft/s
Are you sure? The air resistance is more like proportional to velocity square.
However it does not matter since you do not need to know.
v(0) = 180 * (1- e^0) but anything to the 0 is 1
so
v(0) = 180 (1 - 1) = 0
v(5) = 180(1 - e^-1.15) = 180(1-.317)
= 123
v(10) = 180 (1 - e^-2.3) =180(1-.100)
= 162
a) Well, if we want to find the initial velocity of the sky diver, we need to look at the given equation. It says that the downward velocity, v(t), is given by v(t) = 180(1 − e^(-0.23t)).
Now, when time t is 0, let's see what happens. We plug in 0 for t:
v(0) = 180(1 − e^(-0.23*0))
= 180(1 − e^0)
= 180(1 − 1)
= 180(0)
= 0 ft/s
So, the initial velocity of the sky diver is 0 ft/s. Looks like she's not going anywhere in the beginning!
b) Now let's find the velocity after 5 seconds and after 10 seconds.
For 5 seconds:
v(5) = 180(1 − e^(-0.23*5))
≈ 180(1 − e^(-1.15))
≈ 180(1 − 0.316640)
≈ 180(0.683360)
≈ 123 ft/s (rounded to one decimal place)
For 10 seconds:
v(10) = 180(1 − e^(-0.23*10))
≈ 180(1 − e^(-2.3))
≈ 180(1 − 0.091578)
≈ 180(0.908422)
≈ 163 ft/s (rounded to one decimal place)
So, after 5 seconds, the velocity is approximately 123 ft/s, and after 10 seconds, the velocity is approximately 163 ft/s. Time flies when you're plummeting towards the ground, huh?
(a) To find the initial velocity of the sky diver, we can substitute t = 0 into the given equation for v(t).
v(0) = 180(1 − e^(−0.23(0)))
= 180(1 − e^0)
= 180(1 − 1)
= 180(0)
= 0 ft/s
Therefore, the initial velocity of the sky diver is 0 ft/s.
(b) To find the velocity after 5 seconds, we can substitute t = 5 into the given equation for v(t).
v(5) = 180(1 − e^(−0.23(5)))
= 180(1 − e^(−1.15))
≈ 180(1 − 0.3152)
≈ 180(0.6848)
≈ 123.26 ft/s
After 5 seconds, the velocity of the sky diver is approximately 123.26 ft/s.
To find the velocity after 10 seconds, we can substitute t = 10 into the given equation for v(t).
v(10) = 180(1 − e^(−0.23(10)))
= 180(1 − e^(−2.3))
≈ 180(1 − 0.0924)
≈ 180(0.9076)
≈ 163.16 ft/s
After 10 seconds, the velocity of the sky diver is approximately 163.16 ft/s.
To find the initial velocity of the skydiver, we need to look at the given equation for velocity at time t:
v(t) = 180(1 − e^(-0.23t))
When t = 0, the initial time, we can plug this value into the equation to get the initial velocity:
v(0) = 180(1 − e^(-0.23 * 0))
= 180(1 − e^0)
= 180(1 − 1)
= 180(0)
= 0 ft/s
Therefore, the initial velocity of the skydiver is 0 ft/s.
To find the velocity after 5 seconds, we can substitute t = 5 into the velocity equation:
v(5) = 180(1 − e^(-0.23 * 5))
≈ 180(1 − e^(-1.15))
≈ 180(1 − 0.3157)
≈ 180(0.6843)
≈ 123.4 ft/s (rounded to one decimal place)
Therefore, the velocity after 5 seconds is approximately 123.4 ft/s.
Similarly, to find the velocity after 10 seconds, we substitute t = 10 into the velocity equation:
v(10) = 180(1 − e^(-0.23 * 10))
≈ 180(1 − e^(-2.3))
≈ 180(1 − 0.0956)
≈ 180(0.9044)
≈ 162.8 ft/s (rounded to one decimal place)
Therefore, the velocity after 10 seconds is approximately 162.8 ft/s.