# Velocity and acceleration and drag

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 ?

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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?

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

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

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?

i dunno how

Set v = 0. What do you get?