Question

Is each situation a linear or a nonlinear model?

.
The stopping distance for a car is given by the equation , where D is the distance a car travels once the brakes are applied and v is the car’s velocity.

2.
A home’s value is given by the equation y = 154,000(1.04)x, where y is the value of the house and x is the number of years since the home was purchased.

Answer 1

are you sure that this is the full question (without any information missing) ?

Answer 2

The first situation is missing a formula. The situation 2 we have:
y = 154000*(1.04)^x, right?
Think about the graph of this function... is it linear? (i.e. a line)

Answer 3

If you know enought physics, we can find a equation of the distance D in finction of v:
we have from the energy conservation theory that:
(m*v^2)/2 = m*g*u*D, where m is mass of the car, g is gravitational acceleration, u is the kinect friction coefficient.
D = (v^2)/(2*m*g*u)
From that, if we plot this function, will the be a line? (i.e. linear)?

Answer 4

I usually see cases where the velocity function is just a derivative, so I would have all the rights to assume that #1 's function is not linear...

or it would usually not be linear

Answer 5

If you know enought physics, we can find a equation of the distance D in finction of v:
we have from the energy conservation theory that:
(m*v^2)/2 = m*g*u*D, where m is mass of the car, g is gravitational acceleration, u is the kinect friction coefficient.
D = (v^2)/(2*g*u)
From that, if we plot this function, will the be a line? (i.e. linear)?

Answer 6

You are right @SolomonZelman , as we can see:
\[m\frac{ dv }{dt } = -mg\mu\]\[\frac{ dv }{ dx }\frac{ dx }{ dt } = -g\mu\]\[v \frac{ dv }{ dx } = -g\mu\]\[\int\limits_{v}^{0}vdv = -\int\limits_{0}^{D}g\mu dx\]\[\frac{ v^2 }{ 2 } = g\mu D\]\[D = \frac{ v^2 }{ 2g\mu }\]
As i found above using energy