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Velocity and acceleration and drag

  • #1
This is probably a maths question which im struggling with
the question states that
drag is proportional to the square of the velocity
D = kv^2

And there is a linear relationship between the square of the velocity and the acceleration
dv/dt = - 0.0154 v^2 + 0.402827

assume the mass of the object is 700 kg

F = ma
F = 700 dv/dt
and the force = propulsion- drag = P - D
P - D = 700 dv/dt
dv/ dt = ( P - D) / 700




with this data below how do i find the drag and the propulsion

v^2 (m/s)__________( dv/dt) , ms/s/s
0.6241_____________0.39
5.0176_____________0.32
11.2225____________0.23
16.81______________0.15
20.7025____________0.09
23.2324____________0.05



I tried putting the linear equation in the Newton second law
dv/ dt = ( P - D) / 700
dv/dt = - 0.0154 v^2 + 0.402827
so , ( P - D) / 700 = - 0.0154 v^2 + 0.402827
but how do i find the P and D ?
 

Answers and Replies

  • #2
6,054
390
Are the tabular data simply what you can get from the linear equation? I have tried two rows, and they fit. So I think they can be ignored.

With ( P - D) / 700 = - 0.0154 v^2 + 0.402827, what do you get by multiplying both sides by 700? What form must the drag term have? What else is there?
 
  • #3
when multiply both sides with 700 it becomes
P-D = 281.979 - 10.78v^2
well since D = kv^2
it becomes

P - kv^2 = 281.979 - 10.78v^2
 
  • #4
my teacher kept mentioning the drag is proportion to the velocity , and she told us to find the linear relationship between acceleration and the velocity, and we can find the constant of the drag. but i dont know what she meant by that
 
  • #5
6,054
390
The drag is proportional to the SQUARE of the velocity, at least in this problem. Using the final equation you got, can you determine P and k?
 
  • #6
i dunno how
 
  • #7
6,054
390
Set v = 0. What do you get?
 

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