# Terminal speed

• Nov 3rd 2010, 10:39 AM
Darkprince
Terminal speed
A racing car of mass m travels along a straight road. While travelling, the car is subject to a constant frictional resistance force of magnitude a*m (a, a positive constant) and an air resistance force of magnitude k*(v^2), where v is its speed. The car engine can provide a constant propelling force of magnitude b*m (b a positive constant), which can bring the racing car to a terminal speed u'

a) show that k=m*(b-a)/(u')^2

b) hence show that du/dt = (b-a) (1-(u/u')^2)

c) the car starts from rest at time t=0. Show that the car reaches speed 0.5u' at time t=t1/2,
where t1/2=(u'*ln3)/2(b-a)

d) at time t=t1/2, the engine is switched off. Show that the car comes to rest at time
t1/2 + (u'/sqrt(a*(b-a)))*(tan^-1(sqrt(b-a)/4a))

I have done the first three parts. I have done the first part by considering what is the force balance on the car while is travelling with constant speed v'. I have done the second part by considering that F=m*a => a = f/m I have done also the third part by integrating from 0 to v'/2 of (1/(dv/dt)) dv
I have stucked at part d. I realize that when the engine is switched off the propelling force of the engine stops to act on the car and then we only have the air resistance force and the frictional resistance, two forces that will decrease the speed of the car until the car comes to rest.

Any help would be appreciated. Thanks in advance.

Kind regards!
• Nov 3rd 2010, 12:07 PM
Darkprince
Also, we know that m*(dv/dt)= -m*a-k*v^2 when the engine of the car is switched off. I have tried to solve in respect with t, but that doesn't seem to lead me to the correct answer. Any help would be appreciated! Thanks in advance!
• Nov 4th 2010, 12:42 AM
CaptainBlack
Quote:

Originally Posted by Darkprince
Also, we know that m*(dv/dt)= -m*a-k*v^2 when the engine of the car is switched off. I have tried to solve in respect with t, but that doesn't seem to lead me to the correct answer. Any help would be appreciated! Thanks in advance!

This is of varuiables seperable type, so:

$\dfrac{m}{-am-kv^2}\dfrac{dv}{dt}=1$

hence:

$\displaystyle \int_{v=v(t_{1/2})}^{v(t)} \dfrac{m}{-am-kv^2} \; dv=\int_{\tau=t_{1/2}}^t \;d\tau$

CB
• Nov 4th 2010, 09:05 AM
Darkprince
Than you very much!