# Homework Help: Finding approx. minimum value for 'v'

1. Aug 3, 2012

### xiphoid

1. The problem statement, all variables and given/known data
A 2m wide truck is moving with a uniform speed v=8m/s along a straight horizontal road a pedestrian starts to cross the road with a uniform speed u when the truck is 4m away from him minium value of v?

2. Relevant equations
The equations of motion in a straight line, and the equation of relative velocity, distance....

3. The attempt at a solution
since it is uniform speed, here, acceleration i suppose should be zero. Even after drawing a diagram, couldn't get help from it!
I doubt if this is really solvable using the above mentioned equations that I have so far come across

2. Aug 3, 2012

### dharavsolanki

Minimum implies there is a range on the values.

What is the condition when the minimum happens. What if you are crossing the road with less than the minimum speed?

Imagine physical scenarios :-D

3. Aug 3, 2012

### xiphoid

What do you want me exactly to do?
As I had posted in the question itself, i had drewed the diagram, and perhaps imagine every incident that can take place in my mind,
but
any hints for getting the answer?

4. Aug 3, 2012

### voko

Minimum speed for a surviving pedestrian.

5. Aug 3, 2012

### CWatters

Work out how long the truck takes to cover the remaining 4m. The pedestrian has the same time to get out of the way (eg to travel >2m).

6. Aug 4, 2012

### xiphoid

I have only the intial velocity of the truck and that of the paedestrian, no final velocity!
How can I manage to calculate this without the final velocity?
Assume it to be 0?

7. Aug 4, 2012

### willem2

If the truck has a uniform speed, the final velocity, as well as the velocity at any other time is 8 m/s.

I suppose you're really supposed to find the minimum of the velocity of the pedestrian 'u', since the speed of the truck is already given.

8. Aug 4, 2012

### CWatters

xiphoid - The question is simpler than you think.

Both truck and pedestrian are travelling at a uniform/constant velocity. The truck doesn't brake! The pedestrian doesn't see it coming and walks at a constant velocity.

You have to find the minimium velocity that the pedestrian could be walking at yet still escape being hit. Not the minimium velocity he achieves after being hit or anything as complicated as that. If he runs faster than Vmin he escapes. If he walks slower than Vmin, he's still in the way when the truck arrives and gets hit.

Last edited: Aug 4, 2012
9. Aug 7, 2012

### xiphoid

Ok... so the time I managed to calculate is 0.5s, what next?
should i use the time in any of the equations of the motion, here the acceleration is 0.

10. Aug 7, 2012

### CWatters

Well then that's how long he has to get out of the way. If he only manages to cover 1m in that time the truck will hit him. If he covers 3m in that time the truck will pass behind him. What's the minimium distance he has to cover in that time?

Minimum velocity = minimum distance / time.

11. Aug 8, 2012

### xiphoid

the answer is 8m/s?

12. Aug 8, 2012

### CWatters

2m in 0.5 seconds = ?

13. Aug 8, 2012

### xiphoid

ok... so the answer is 4m/s, but that doesn't appears in the options given to me????

14. Aug 8, 2012

### dharavsolanki

See, this is a VERY formulae based approach. We don't start with formula and apply it to situations, we draw/define situations, find what we need and THEN find out situations. I guess you started with v = u + at and then digressed into finding v/t.

THIS is why i suggested that you take a look at How To Solve It by Poloya.

Anyways, you have been very persistent in this question, so I am uploading a detailed solution.

I have attached an image here that might give you some idea.

#### Attached Files:

• ###### Vasu's Reply.gif
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15. Aug 8, 2012

### dharavsolanki

That sucks. What are the options given to you? And also, what is the source of the question?

16. Aug 8, 2012

### xiphoid

Actually, I would like to quote this to everyone who has helped me including you, Dharavbhai, that my first answer to the question, the moment I saw was 4m/s but it was not given in the options and plus, somehow i forgot to write that in the calculations done list as well!
the correct answer given is 3.567m/s, dont know how did they got the answer in this much detail!
And the source is Pradeep for NCERT

17. Aug 8, 2012

### dharavsolanki

Then, was it the exact question?

Also, need more proof on why to stay away from Pradeep?

18. Aug 8, 2012

### CWatters

If both truck and pedestrian go at uniform velocity I can't think of any possible way the pedestrian could avoid being hit by walking SLOWER than 4m/s.

If the truck was driving say 0.5m away from the sidewalk then the pedestrian would have to go faster than 4m/s.

The ONLY way I can see for him to escape by going slower than 4m/s is if the truck slows down or swerves onto the sidewalk.

Could you type in the problem _exactly_ as set word for word including any diagrams? The answer is either wrong or we have totally missunderstood the question.

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