Motion Problem: A car travels along a horizontal road at....

F

f1player

Junior Member
Joined
Feb 25, 2005
Messages
59
The problem is as follows:

A car travels along a horizontal road at 30m/sec. The engine is stopped and the car allowed to coast to rest. Frictional resistance is 1/10 the weight of the car. Air resistance is v/20 Newton's for each kilogram of mass. Find the distance before the car comes to rest.

I used Net Force = ma to get an expression for a in terms of v:

0.1mg + vm/20 = ma

So: 0.98m + vm/20 = ma

a = 0.05v + 0.98

I know that a = dv/dt and that I can invert that so that dt/dv = 1/(0.05v + 0.98)
When I integrate this i get: t = 20ln(0.05v + 0.98) + c

This is where I get lost. At this point is it true to say that when t = 0, v =30 ??
Otherwise, how else would you find the constant of integration??
 
net force = frictional resistance + air resistance

ma = -(mg/10 + mv/20)

a = -(2g + v)/20

now ... when v = 0 (the car has coasted to a stop) note that a acceleration is still
-g/10 due to frictional resistance. Rolling frictional resistance should also be a function of the car's velocity ... something is missing here.
 
So you're saying that there's information missing in the question??

I've written out the exact question there as it was given to me, so as far as I know all the info is there for the question to be solved.
 
ok ... let's solve it as given.

ma = -(mg/10 + mv/20)

a = -(2g + v)/20

dv/dt = -(2g + v)/20

dv/(2g + v) = -1/20 dt

ln(2g + v) = -t/20 + C<sub>1</sub>

2g + v = C<sub>2</sub>e<sup>-t/20</sup>

v = C<sub>2</sub>e<sup>-t/20</sup> - 2g

at t = 0, v = 30 m/s ...

30 = C<sub>2</sub>e<sup>0</sup> - 2g

C<sub>2</sub> = 30 + 2g

v = (30 + 2g)e<sup>-t/20</sup> - 2g

now ... setting v = 0 to find the time t when the car stops ...

0 = (30 + 2g)e<sup>-t/20</sup> - 2g

(30 + 2g)e<sup>-t/20</sup> = 2g

e<sup>-t/20</sup> = 2g/(30 + 2g)

e<sup>-t/20</sup> = g/(15 + g)

-t/20 = ln[g/(15 + g)]

t = (20)ln[(15 + g)/g] = approx 18.569... sec

integrating velocity from t = 0 to t = 18.569... yields a distance of approx 236 m.
 
Ok, here's what I did:

F = -(0.1mg + vm/20) = ma

F = -0.1mg - 0.05vm = ma

So: a = -0.1g - 0.05 v

a = -0.98 - 0.05 v (Using g as 9.8)

dv/dt = -0.98 - 0.05v

dt/dv = 1/(-0.98 - 0.05v)

So: t = 20ln(-0.98 - 0.05v) + c (When t = 0, v = 30)

Now at this point I can't find c, because the ln value is negative, so it is a complex number. How do I proceed from here?? Maybe I got my negatives mixed up??
 
Ok, doesn't matter. I can do it now

I've gotten the answer of 236.04 m, so about 236 m

Thanks for all the help anyway
 
You must log in or register to reply here.
Top