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Math Quiz Thread

I think you are right, I under defined the problem, and bomb was right. An infinite number of solutions.

Live and learn. (y)
 
Not sure why I sat hear and read the whole thread. I didn't understand any of it.
 
Gospel said:
Not sure why I sat hear and read the whole thread. I didn't understand any of it.
Click to expand...
The problems were posed as a circle as an obfuscation.

Two cars are traveling in the same direction on a long straight road. They are 100 meters apart.

The lead car is going 1 meter per second,
The trailing car is going 2 meters per second.
How long before the trailing car catches up with the leading car?

When the distance traveled by the trailing car equals the distance traveled by the lead car plus 100 meets the cars will be abreast.

Distance = velocity times time, v*t. Go 50 miles per hour for 1 hour and you go 50 miles.
v1 = 2 meters per second
v2 = 1 meters per second

so v1*t = 100 + v2*t, and solve for t.

Applying algebra

v1*t = 100 + v2*t
v1*t - v2*t = 100
t*(v1-v2) = 100
t = 100/(v1-v2)
 
Are you Starfleet Academy material?

You are in space away from appreciable influence of gravity and you need to launch a probe in a straight line.

Th velocity time profile is exponential that accelerates to a target velocity above the space ship velocity. The acceleration has to be limited to protect equipment in the probe,

Given:
probe mass = 1000kg
maximum acceleration = 10 g's
desired probe velocity = 100m/s above ship velocity

Determine probe:
velocity(t)
acceleration(t)
force(t) in Newtons
engine_power(t) in Watts
total energy consumed in Joules
 
steve_bank said:
Are you Starfleet Academy material?

You are in space away from appreciable influence of gravity and you need to launch a probe in a straight line.

Th velocity time profile is exponential that accelerates to a target velocity above the space ship velocity. The acceleration has to be limited to protect equipment in the probe,

Given:
probe mass = 1000kg
maximum acceleration = 10 g's
desired probe velocity = 100m/s above ship velocity

Determine probe:
velocity(t)
acceleration(t)
force(t) in Newtons
engine_power(t) in Watts
total energy consumed in Joules
Click to expand...
Huh?

You have the first two values already. And note that you set a maximum acceleration, not a minimum, force can be anything up to 98,100 Newtons. Watts is a simple conversion from Newtons.

Anyway, how about an old one with a new twist?

Original: A guy is driving along and is pulled over by a cop for running a red light. He explains he was going so fast that the red (700nm) looked like green (550nm). The speed limit is 100 km/hr, the fine is $50 + $1 per km/hr over the limit. What's the fine?

New twist: He's testing a device that alters local lightspeed. What lightspeed is he experiencing?
 
Feel free to show a solution. I do not think you are right.

What equations do you enter into the probe guidance system?
 
First exponential growth.


Exponential growth asymptotically approaching a value of k vs time
y(t) = k(1 – e(-t/tau))

tau is called the time constant. At t = tau e^(-t/tau) = 1/e = .367 and y = .632*k approximately. Tau fixes y versus t at a point.

So for the probe v(t) = v_target(1-e(-t/tau)) and the problem is to determine tau. For the problem there is no time requirement to reach the target velocity, there is an acceleration limit. To solve for tau take the first derivative, apply the acceleration constraint and solve for tau.

V'(t) = a(t) acceleration.

An interesting property of exponentials is that the derivative and integral of an exponential is an exponential.

Y = e^u y' = e^(u)du
y = e^x y' = e^x
y = e^(kx) y' = ke^(kx)
y = e^x integral e^x b to a is e^a – e^b

Handy to remember considering the prevalence of exponential functions..

v(t) = v_target(1-e(-t/tau))
a(t) =(-v_max/-tau)^(-t/tau)
At t = 0 a(t) = a_max and e^(-t/tau) = 1, so
a_max = v_max/tau
tau = v_max/a_max.

Substituting back in
a(t) = a_max*e^(-t/tau)

v(t) = v_target*(1- e^(-t/tau)) m/s
a(t) = a_nax*e^(-t/1tau) m/s^2
F(t) = m_drone * a(t) Newtons

The work integral F*distance is solved numerically

Code: 
option explicit

Sub Main

Dim Doc As Object
Dim cell as object
Dim Sheet As Object
Doc = ThisComponent
Sheet = Doc.Sheets (0)
Cell = Sheet.getCellByPosition(0,0)
cell.string = "RUNNING"


dim n,i as integer
dim _e,a_max,v_target,tau,dt,tmax,m_drone,_t,g,ds as double
_e = 2.1782818
g = 9.8
a_max = 10. * g
m_drone = 1000. 'kg
v_target = 100. 'm/s
dt = .1
tau = v_target/a_max
tmax = 10 * tau  '10 time constnts seconds
n = tmax/dt
_t = 0.
dim F(n),a(n),t(n),w(n),E(n),v(n) as double

for i = 0 to n
    t(i) = _t
    _t = _t + dt
    v(i) = v_target * ( 1 - _e^(-t(i)/tau))
    a(i) = a_max * (_e^(-t(i)/tau))/g
    F(i) = m_drone * a(i) 'F=ma
    ds = v(i) * dt
    E(i) = F(i) * ds  'joules = force * distance
    W(i) = (E(i)/dt)/1e3  'kilowatts
next i

Cell = Sheet.getCellByPosition(1,0)
cell.string = "TIME"
Cell = Sheet.getCellByPosition(2,0)
cell.string = "VELOCITY"
Cell = Sheet.getCellByPosition(3,0)
cell.string = "ACCEL"
Cell = Sheet.getCellByPosition(4,0)
cell.string = "FORCE"
Cell = Sheet.getCellByPosition(5,0)
cell.string = "POWER"

for i = 0 to n

    Cell = Sheet.getCellByPosition(1,i+1)
    cell.value = t(i)
    Cell = Sheet.getCellByPosition(2,i+1)
    cell.value = v(i)
    Cell = Sheet.getCellByPosition(3,i+1)
    cell.value = a(i)
    Cell = Sheet.getCellByPosition(4,i+1)
    cell.value = F(i)
    Cell = Sheet.getCellByPosition(5,i+1)
    cell.value = W(i)
next i

Cell = Sheet.getCellByPosition(0,0)
cell.string = "DONE"

End Sub
 
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