Hyperloop: Drag

$m=15 \ 000 \ \mathrm{kg}$ is the pod's mass. $v_0=300 \ \mathrm{\frac{m}{s}}$ is the pod's initial velocity. $v_f=150 \ \mathrm{\frac{m}{s}}$ is the pod's final velocity. $F_0=320 \ \mathrm{N}$ is the drag force on the pod at $v=v_0$ .

We can find the constant $C$ in $F_d=Cv^2$ by setting $F_d=F_0$ and $v=v_0$ , which gives $C=\frac{F_0}{v_0^2}$ . So the force on the pod is $$F= -F d=-\frac{F 0 v^2}{v 0^2}.$$ By Newton's Second Law, we have $$a=\frac{F}{m}.$$ Substituting in for $F$ and noting that $a=\frac{dv}{dt}$ , we get $$\frac{dv}{dt}=-\frac{F 0 v^2}{v 0^2 m}.$$ This is a separable differential equation, so separating variables and integrating gives us $$\int \frac{1}{v^2}dv=\int \frac{-F 0}{v 0^2 m}dt$$ $$\frac{-1}{v} = \frac{-F 0}{v 0^2 m}t + k,$$ where $k$ is the constant of integration. Plugging in the initial condition of $v(0)=v_0$ gives $$k=\frac{-1}{v 0}.$$ So we have $$\frac{1}{v} = \frac{F 0}{v 0^2 m}t + \frac{1}{v 0}.$$ Now, plugging in $v=v_f$ and solving for $t$ gives $$t=\left(\frac{1}{v f}-\frac{1}{v 0}\right)\left(\frac{v 0^2 m}{F_0}\right)$$ Plugging in the values given, we get $$t=\left(\frac{1}{150 \ \mathrm{\frac{m}{s}}}-\frac{1}{300 \ \mathrm{\frac{m}{s}}}\right)\left(\frac{(300 \ \mathrm{\frac{m}{s}})^2 (15 \ 000 \ \mathrm{kg})}{320 \ \mathrm{N}}\right)$$ $$=\fbox{14 062.5 s}$$

We first find the value of the constant $C$ : $C=\frac{F_d}{v^2}=\frac{320}{300^2}=3.556\times 10^{-3} \, \mathrm{kgm}^{-1}\mathrm{s}^{-4}$

Note that $F_d=ma=m\frac{\mathrm{d}v}{\mathrm{d}t}$ . Now to find the time, we will solve the differential equation: $m\frac{\mathrm{d}v}{\mathrm{d}t} = -Cv^2$ $m\frac{\mathrm{d}v}{v^2} = -C\mathrm{d}t$ $m\int_{300}^{150} \frac{1}{v^2} \, \mathrm{d}v = -\int_0^T C\, \mathrm{d}t$ $m \left( \left.\frac{-1}{v} \right| ^{150}_{300} \right) = -CT$ $m\left(\frac{-1}{150} + \frac{1}{300} \right) = -CT$ $T= \frac{m}{C} \left( \frac{1}{300} \right) = \frac{15000}{3.556\times 10^{-3}}\left( \frac{1}{300} \right) =14060 \, \mathrm{ s}$

We solve for C by manipulating $F_d=Cv^2$ to get $C=\frac{F_1}{v_0^2}$ . We keep in mind that this will be negative as $F_1$ is a resistive force. Next, we attempt to find the acceleration: $ma=Cv^2$ . We rewrite as $m\dot{v}=Cv^2$ and separating variables, we have $m\frac{dv}{v^2}=Cdt$ . Integrating, we get $-\frac{m}{v} = Ct + k$ , where k is the constant of integration. We plug in t=0 and find that $k=-\frac{m}{v_0}$ , and $\frac{m}{v_0}-\frac{m}{v}=Ct$ . Solving for t, we find that $t=\frac{\frac{m}{v_0}-\frac{m}{v}}{C} = 14062.5$ .

First, we can find the value of constant C:

$\mathbf{320 = C\times 9\cdot 10^{4} \leftrightarrow C = \frac{32}{9}\cdot 10^{-3}\cdot \frac{kg}{m}}$

Now we can find the expression of the aceleration as a function of the speed:

$\mathbf{F = \frac{32}{9}\cdot 10^{-3}\times v^{2} = 15\cdot 10^{3}\times (-a)}$

$\mathbf{a = \frac{-32}{135}\cdot 10^{-6}\times v^{2}}$

Now, we can integrate the expression of the instant velocity, defined at the intervals $[0, t]$ and $[v_{0}, v]$ :

$\mathbf{dv = a\times dt}$

$\mathbf{dt = \frac{dv}{a} \leftrightarrow \int_{0}^{t} dt = \int_{v_{0}}^{v} \frac{dv}{a}}$

$\mathbf{t = \frac{-135\cdot 10^{6}}{32}\cdot \int_{v_{0}}^{v} \frac{dv}{v^{2}} = \frac{-135\cdot 10^{6}}{32}\cdot (\frac{-1}{v} + \frac{1}{v_0})}$

Solving for $v_{0} = 300m/s$ and $v = 150m/s$ we find:

$\mathbf{t \cong 14060s}$

To find the amount of time it takes to decelerate, $t$ , we need to find the deceleration magnitude $a$ . Then, if the speed is $v$ , we would have $v = at \implies t = \dfrac {v}{a}.$ But, from Newton's Second Law, $a = \dfrac {F_d}{m}$ . Thus: $t = \dfrac {v}{a} = \dfrac {v}{\dfrac {F_d}{m}},$ where $m$ is the mass. We have $t = \dfrac {vm}{F_d},$ so we have all that we need.

Substitute $F_d = 320 \, \text{N}$ , $v = 300 \, \text{m/s}$ and $m = 15000 \, \text{kg}$ to get $t = \boxed {1.41 \times 10^4 \, \text{s}}.$ Remark : Note that the formula for drag force was not required.

The sum of the forces equals the acceleration, in this case there is only one force so we get the next differential equation : $M\frac{\mathrm{d}v}{\mathrm{d}t} = -C v^2.$ We use separation of variables : $\frac{M}{C} \int \frac{\mathrm{d}v}{v^2} = -t+C_1 \Longrightarrow v(t)= \frac{M}{Ct+C_1}.$ Now, we have : $C= \frac{350}{300^2}$ , and from the initial conditions we get : $v(0)=300\Longleftrightarrow \frac{M}{C_1}=300\Longrightarrow C_1=50.$ Now : $v(t)= 150 \Longrightarrow t= \frac{M-150C_1}{150C} = \frac{28125}{2} = 14062.5.$

First we find C.

$320=C\times300^2\\ C=\frac{4}{1125}$

We know that $F=m a=-C v^2$ .

Substituting constants and rewriting a and v as x-derivatives,

$15000 x''(t)=-\frac{4}{1125}x'(t)^2$

The solution to this is:

$x(t)=4218750 \log \left(t-4218750 c_1\right)+c_2$

Initial conditions at $t=0$ are $x=0$ and $v=300$ . Substituting and solving for unknown constants give:

$c_1=-\frac{1}{300}\\ c_2=-4218750 \log \left(\frac{28125}{2}\right)\\ x(t)=4218750 \log \left(\frac{2 t}{28125}+1\right)\\ x'(t)=\frac{300}{\frac{2 t}{28125}+1}$

We are searching for $t$ when $v=150$ :

$v=x'(t)=\frac{300}{\frac{2 t}{28125}+1}=150\\ t=14062.5$

c=320/〖300〗^2 , c v^2=ma, c v^2=m (dv/dt), C/m dt=dv/v^2 ∫ (t 1)^(t 2)▒〖C/m*dt〗=∫ 300^150▒dv/v^2
thats all :D.

The first information about the f=cv^2 for finding the constant c. Hence putting the appropriate values i.e f=320 and v = 300, we find "c". Next using the assumption m.dv/dt=cv^2 as drag is the only force acting on the system. On further solving, we get m.dv/v^2 = c.dt. On integrating and putting the limit of 'v' from 300 to 150, we get the appropriate answer that is 14062.5 sec.

mass m=15000 kg drag force when pod travelling at 300 m/s=320 N since, $drag force=C v^{2}$ so, $C v^{2}=320$ at v=300 from above equation we can find the value of constant C. now from calculus $a= \frac{dv}{dt}$ , putting $a= \frac{(Cv^{2})}{m}$ and solving the differential equation with constrains we get the required answer.

F=Cv^2 So to find C, the initial conditions of F=320N when v=300m/s can be used, giving C to be about 3.56*10^-3 .

Now this force is the only force on the pod, so is also the net force. {F(net)=ma=m(dv/dt)=-Cv^2} resulting in {- v^(-2) dv = C/m dt}

integrating between 300 and 150 for the velocity, and 0 and T for time: C/m T = 1/150 -1/300

Substituting in the C and m values:

yielding T=1.4*10^4 s

First we solve for the constant $C$ . We're given $F_d=320$ when $v=300$ . Plugging into the drag force equation leads to $320 = C \times 300^2$ , yielding $C = \frac{4}{1125}$ .

Next, we write the equations of motion for the system (plugging $F_d = Cv^2$ into $F = ma$ ): $\frac{dv}{dt} = \frac{C v^2}{m}$ The solution is $\begin{aligned} \frac{1}{v^2} dv &= \frac{C}{m} dt \\ \int \frac{1}{v^2} dv &= \int \frac{C}{m} dt \\ -\frac{1}{v} &= \frac{C t}{m} + C_v \\ \end{aligned}$ where $C_v$ is the constant of integration. We use this equation for the remaining calcuations. Plugging in the given initial conditions ( $t=0, \; v=300$ ) reveals that $C_v = -\frac{1}{300}$ . And finally solving for the unknown $t$ at the desired velocity $v=150$ using our newly found $C_v$ shows that: $t = \frac{28125}{2}$ Which is 14062.5 in decimal.

From Newtons Second Law, $F = \frac{dp}{dt} = m \frac{dv}{dt}$ . We also have $F = Cv^2$ .

We can determine the value of $C$ to be $C = \frac{F_d}{v^2}$

Equating these expressions for the force on the pod, we get $m \frac{dv}{dt} = Cv^2$ , which is a separable ODE in $v$ and $t$ . Solving this equation for $t$ yields $t = \frac{m}{C} \cdot v^{-1} + A$ , where $A$ is arbitrary. Setting $A = 0$ and plugging in the values for $v$ and $C$ we get $t = \boxed{14060.5} \text{ s}$

Acceleration $=\frac{dv}{dt}=\frac{F}{m}$

$320N=Cv^{2}$

$C=\frac{320N}{(300m/s)^{2}}$

Now we equate acceleration to drag force, and $Cv^{2}$ is negative here because the force is applied in the direction opposite to motion, whose direction we are taking to be positive.

$\frac{dv}{dt}=-Cv^{2}$

$-\frac{1}{Cv^{2}} dv=\frac{1}{m} dt$

$\frac{1}{Cv}=\frac{t+c}{m}$

$t+c=\frac{m}{Cv}$

Subbing in the initial values of 300m/s at 0s and the values of C and m,

$c=\frac{15000kg}{\frac{320N}{(300m/s)^{2}} \times 300m/s}=14,062.5s$

Now plugging in values to the equation we have,

$14,062.5s+t=\frac{15000kg}{150C}$

$t=14,062.5s=14,100s (3sf)$