What is the best approach for solving this problem?

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To solve the problem of a 6000 Kg lorry brought to rest from 80 km/h in 2 minutes, the initial speed must be converted to meters per second, resulting in 22.224 m/s. The retarding force can be calculated using Newton's second law after determining the deceleration from the change in momentum over the time interval. For the distance traveled during retardation, using the average velocity method is valid, as it simplifies the calculation. Some participants suggest a more dynamic approach, emphasizing the relationship between impulse and work done rather than relying solely on kinematic equations. The discussion highlights different valid methods for solving the problem, ultimately leading to a consensus on the approach.
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Homework Statement


A 6000 Kg lorry is traveling at 80 km/h on a level road. It is brough to rest with a uniform retardation in 2 minutes. Calculate a) the retarding force and b) the distance traveled during retardation.


Homework Equations


now I know I've got to change the 80km/h into m/s, which is 80 x 0.2778 = 22.224 m/s. for distance traveled its something like S=ut+1/2*-at2. So if someone could explain how to go about doing that question I would be very greatful.
 
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did you calculate the retarding force ?

the second part can be solved using

d = v_{avg} t
where Vavg is the average velocity
 
fable121 said:

Homework Statement


A 6000 Kg lorry is traveling at 80 km/h on a level road. It is brough to rest with a uniform retardation in 2 minutes. Calculate a) the retarding force and b) the distance traveled during retardation.


Homework Equations


now I know I've got to change the 80km/h into m/s, which is 80 x 0.2778 = 22.224 m/s. for distance traveled its something like S=ut+1/2*-at2. So if someone could explain how to go about doing that question I would be very greatful.

You need to find the deceleration, which you can do with the information in the question. Do you know the formula for that? You can then find the net force by using Newton's second law.

Your equation for distance will work fine, if you get all the variables right (including the signs).
 
general method:

since you know the change in momentum of the truck (er, lorry), you know the impulse that was supplied. Since you also know the time interval over which this impulse was supplied, you can calculate the retarding force (assuming it is uniform, which is stated in the problem).

Furthemore, since you know the change in kinetic energy of the truck, you know the work done on it, and can therefore calculate the distance over which this work was done (since you know the retarding force). This strikes me as simpler/more elegant than resorting to kinematics to find the distance travelled, but maybe others would disagree.
 
Ok I think I can do it now, thanks for the help all :)
 
stunner5000pt said:
did you calculate the retarding force ?

the second part can be solved using

d = v_{avg} t
where Vavg is the average velocity

Why use the average velocity??
 
fable121 said:
Ok I think I can do it now, thanks for the help all :)



no problem. feel free to post your work here if you would like it checked over, or if you run into trouble.

EDIT: I added a quote to make it clear who I was replying to.
 
hage567 said:
Why use the average velocity??

Good question. I don't know why he suggested that. I think it's valid though, right? Mean value theorem?

T = 2 \ \textrm{min}

\Delta s = \int_0^T v(t) \, dt = \left( \frac{1}{T} \int_0^T v(t) \, dt \right) T \equiv v_{\textrm{avg}} T
 
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cepheid said:
Good question. I don't know why he suggested that. I think it's valid though, right? Mean value theorem?

T = 2 \ \textrm{min}

\Delta s = \int_0^T v(t) \, dt = \left( \frac{1}{T} \int_0^T v(t) \, dt \right) T \equiv v_{\textrm{avg}} T

method has worked for me in the past ...

where would it fail, though?
 
  • #10
stunner5000pt said:
method has worked for me in the past ...

where would it fail, though?

hmmm? Oh, I never said it would fail. I just had the opinion that a completely dynamical solution to the problem was the best approach, without resorting to kinematics. After taking the time to think about it (post #8), though, I think it's a perfectly valid method.

I'm not sure what hage's specific objection was, that's why I tried to verify that the method made sense in my last post.
 
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