# Velocity-Time relationship for acceleration/deceleration

1. Nov 13, 2009

### grey

Not sure if the question is sophisticated enough to be posted here, but here goes...

I am trying to work out the velocity-time relationship for a vehicle under certain conditions. The 'certain conditions' are:
1. deceleration due to aerodynamic drag and rolling friction
2. acceleration due to an engine less drag and friction

now, i will address the deceleration first. I tried to do model this in two ways:
a) Equilibrium of forces giving: -D - f = ma (where D=drag, f=friction). putting in the known values, i get the value of acceleration in terms of velocity as:
a = -0.00075v2-0.2943
i know that a=dv/dt, so re-arranging, i get: dt=dv/a as a=f(v)
Integrating this, i got:
v = 19.8 x tan (((t1-t)/67.3) - tan-1(v/19.8))
now, QUESTION: does this formulation make sense? and second, would the angle be in radians or degrees?

i put in values both in degrees and radians, but found the deceleration to be quite fast with radians and very slow with degrees. By fast and slow, i mean not in sync with practice. For example:
with radians, the vehicle goes from 10 m/s to 5m/s in 15s with a fairly constant gradient. Does it make sense?

b) with the basic conservation of energy equation, starting from:
KE1 - (Pf + PD)xt = KE2
Putting the values in, i got this relationship b/w time-velocity:
t = 666.7 x (100 - v2)/(v x (v2+392.4))
the deceleration (v-t) curve i got was more moderate, with a markedly decreasing gradient
QUESTION: How do these two methods compare?
What am i doing wrong?
What else should i be doing?

2. Nov 22, 2009

### bluelava0207

Well, both a) and b) look fine. Whichever way you choose, you should realize that you have a nonlinear differential equation. I don't myself know how to analytically solve those (get an equation), though I can get MATLAB to simulate it and plot the results.

What is the initial speed?

Oh, and radians. The radian is the fundamental "unit" for angles. Degrees are derived from radians. Calculators use radians unless you somehow tell it you want degrees.

3. Nov 23, 2009

### tyroman

4. Nov 23, 2009

### grey

hey tyroman and bluelave0207,

thanks a lot for your replies. I managed to solve the DE and the result was pretty much similar to the one i posted above: v = 19.8 x tan (((t1-t)/67.3) - tan-1(v/19.8)). Confirmed it in some book as well.

However, now i am stuck at the 2nd case:
"2. acceleration due to an engine less drag and friction"
In the solutions to the DE with the added term of engine torque (from the Newton's second law formulation), I haven't been able to work it out properly because there is always some sort of thing that shouldn't be there like the natural log of a negative number or square root of a negative number. Any ideas? I can post my attempts if someone want to have a look...