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Mechanics Problem, Cannot understand

  • Thread starter mathwurkz
  • Start date
41
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Here is the problem.

A car starts moving rectilinearly, first with acceleration [tex] \omega = 5.0 [/tex] m/s^2 (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate [tex] \omega [/tex], comes to a stop. The total time of motion equals [tex] \tau = 25 [/tex] s. The average velocity during that time is equal to [tex] v = 72 [/tex] km per hour. How long does the car move uniformly?

If I did not have to deal with the average velocity, then I can figure out that the car will reach a maximum speed of 50 m/s for an instant before it has to decelerate in order to have a 25 s trip. But I think this is where I have a weakness when it comes to average velocities. How do you put it together with acceleration? The answer at the back of my book gives a result of 15 s. I'd appreciate any help getting me pointed in the right direction.
 

Answers and Replies

Fermat
Homework Helper
872
1
create equations of motion for all three stages of the journey.

For the distance travelled and for the velocities at the end-points of each stage.

Remember that the distance travelled and time taken are the same for the 1st and 3rd stages.

Edit: I got 15s too.
 
38
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Pretty much what Fermat said:

Find the distance traveled during the first and second states. Note the distance and time displaced during the 3rd stage is the same as the first. So 2*x1+x2 = 500 m and 2t1 + t2 = 25 s. Find the equations for x1 and x2 in terms of t1 and t2, then solve the system of equations.
 
41
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Thank you I finally solved it. However, what still puzzles me is what the answer at the back gives. I don't understand their interpretation. Then again I don't know if it is even correct since I think there would be a neagtive square root.
What the book gives me:

[tex] \Delta t = \tau \sqrt{\frac{1-4v}{\omega \tau}} = 15 s[/tex]
 

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