A sky diver with a mass of 70 kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be FD =k
V2, where k = 0.25 N*s^2 / m^2.
a. Determine the maximum speed of free fall for the sky diver and the speed reached after 100 m of fall.
b. Plot the speed of the sky diver as a function of time and as a function of distance fallen.
a. Determine the maximum speed of free fall for the sky diver and the speed reached after 100 m of fall.
b. Plot the speed of the sky diver as a function of time and as a function of distance fallen.
1 answer:
Answer:
The correct answer is
The maximum speed of free fall for the sky diver is 52.41 m/s
and the speed reached after 100 m is 37.44 m/s
Explanation:
From Force = mass × acceleration
Aerodynamic drag force = FD = k·V², where k = 0.25 N·s²/m²
The maximum speed, v, of free fall where m·a = 0 at terminal velocity
Mass × Gravity - Aerodynamic drag force = ma = 0
∴ 70 kg × 9.81 m/s² - 0.25 N·s²/m²× v² =0
Therefore v = 52.41 m/s
The speed after 100 m of free fall is given by
V =
Substituting we have, where y - y₀ = 100 then v = 37.44 m/s
b) To plot the relation we have
Speed to time relation as v or s(t) =
Where S(t) is he speed at time t
and the relation between time and distance = v × t
Please see the attached graphs
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Answer:
V_t=6 = 32 m/s
Explanation:
The origin is at point 0 with the positive motion to the right
The instantaneous acceleration is change of velocity measured at infinitesimal interval of time, so the expression for instantaneous acceleration is:
a=dv/dt
From here we can express dv as:
dv = a dt
Replace a by 2t — 1
dv = (2t — 1) dt
Integrate both sides of equation
v=t
a=t_0
putting these value in integral
<em>v-v_0=(t^2-t)-(t_0^2-t_0)</em>
We know that v_0 = 2 at t_0 = 0, so we'll replace t_0 and v_0 by their values
v — 2 = (t^2 — t) — (0^2 — 0)
From here we can write the expression for v as:
v_t=6=6^2-6+2 (1)
So the velocity at t = 6 s is:
v_t=6 = 32 m/s
V_t=6 = 32 m/s
In order to determine the total distance travelled, we must check how maw times the particle has changed its direction, i.e. how many times its speed was equal to zero
To do that, we'll just replace v by 0 in expression (1)
0 = t^2 — t + 2
The roots of the quadratic equation are:
t_1/2=1± √(1^2-4*2*1)/2
Since 1^2-4*2*1 < 0, the quadratic equation have no real roots, so we can say that the velocity is always positive, i.e. to the right
Now that we have all the details, we can correctly draw the path of the particle
We can see from the sketch that the total distance traveled is:
s^T=Δs_0-1
s^T=| s_1 - s_0 |
Replace s_0 by its value
s^T=| s_1 - 1 | (2)
In order to determine the position of particle at t = 6 s, we'll need to determine the expression for s as function of time
Since we have already wrote expression for v as function of time (step 2), we'll use expression to get the expression for s
v= ds/dt
Multiply both sides of equation by dt
v dt = ds
Replace v by expression (1)
(t^2 — t + 2) dt = ds
Integrate both sides of equation
t=s
b=(s=0)
x=(t^2 — t + 2)
dx=ds
putting these value in integral
(t^3/3-t^2/2+t)-(t_0^3/3-t_0^2/2+t_0)= s-s_0
Since s = 1 m at t = 0, and we want to determine the position s at t = 6, we'll replace so by 1, t_0 by 0 and t by 6
(6^3/3-6^2/2+6)-(0^3/3-0^2/2+0)=s_t=6-1
