1. ## kinematics/calculus

hi, here is a paraphased question i am having trouble with:

A vehicle's mass (including passenger) = 400 kg
after travelling 400 m the car stops accelerating. At this instant, the car's brakes are applied, and in addition, a small parachute opens at the rear to slow the car down.
The retarding force applied by the brakes including friction is 5000 N. The retarding force due to the parachute is 0.5v^2 N where v m/s is the velocity of the car x meters beyond the 400 m mark.
a) if a m/s^2 is the acceleration of the car during the retardation stage, write the equation of motion for the dragster during this stage:
.5v^2 + 5000 = 400a
or (1/2)v^2 = 400a - 5000
is this right?

b) By choosing an appropriate derivative form for acceleration, show that the differential equation relating v to x is:
dv/dx = -(10^4 + v^2)/(800v)

Now, im thinking that there is a form for acceleration which goes:
d((1/2)v^2)/dx = a
which looks awfully similar to my (1/2v^2)
but im not too sure where to take it to show that equation.

Any help?

2. Originally Posted by scorpion007
hi, here is a paraphased question i am having trouble with:

A vehicle's mass (including passenger) = 400 kg
after travelling 400 m the car stops accelerating. At this instant, the car's brakes are applied, and in addition, a small parachute opens at the rear to slow the car down.
The retarding force applied by the brakes including friction is 5000 N. The retarding force due to the parachute is 0.5v^2 N where v m/s is the velocity of the car x meters beyond the 400 m mark.
a) if a m/s^2 is the acceleration of the car during the retardation stage, write the equation of motion for the dragster during this stage:
.5v^2 + 5000 = 400a
or (1/2)v^2 = 400a - 5000
is this right?
This looks OK (accept the acceleration is in the opposite direction to the
velocity and in the direction of decreasing x so we ned a - sign here), but
given what the next part of this question is I would arrange it differently,
like:

a = -(v^2 + 10000)/800.

b) By choosing an appropriate derivative form for acceleration, show that the differential equation relating v to x is:
dv/dx = -(10^4 + v^2)/(800v)

Now, im thinking that there is a form for acceleration which goes:
d((1/2)v^2)/dx = a
which looks awfully similar to my (1/2v^2)
but im not too sure where to take it to show that equation.

Any help?

a = -(v^2 + 10000)/800,

or:

dv/dt = -(v^2 + 10000)/800

but dv/dt = dv/dx dx/dt = v dv/dx, so

v dv/dx = -(v^2 + 10000)/800,

rearranging:

dv/dx = -(v^2 + 10000)/(800v)

RonL

3. Ummm....

CaptainBlack is correct, I'm just not sure where he's going with it.

I would use

a = v*(dv/dx) = [(1/2)d(v^2)/dx]

as you were originally thinking.

Then:

a = -(v^2 + 10000)/800

v(dv/dx) = -(v^2 + 10000)/800

dv/dx = -(v^2 + 10000)/(800v)

-Dan

4. Originally Posted by topsquark
Ummm....

CaptainBlack is correct, I'm just not sure where he's going with it.
What I am trying to do is start from where scorpion007 is and reach
what he is asked to derive (if you ignore that I had typed da/dx
throughout where I meant dv/dt )

RonL

5. Originally Posted by CaptainBlack
What I am trying to do is start from where scorpion007 is and reach
what he is asked to derive (if you ignore that I had typed da/dx
throughout where I meant dv/dt )

RonL
That was what I was having the problem understanding.

-Dan

6. Originally Posted by topsquark
That was what I was having the problem understanding.

-Dan
What is particulary annoying about that is that on the paper next to the
computer where I outlined what I am doing it does read dv/dt

RonL

7. thanks guys, i get it now