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Having trouble with some simple physic questions.

  • Thread starter Mr. Fish
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  • #1
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1. The problem statement, variables and given/known data
A tourist being chased by an angry bear is running in straight line toward his car at a speed of 4.0m/s. The car is a distance d away. The bear is 26m behind the tourist and running at 6.0m/s. The tourist reaches the car safely. What is the maximum possible value for d?


Homework Equations


V = Vo + at
X = 1/2(Vo + V)t
X = Vo * t + 1/2at^2
V^2 = Vo^2 + 2ax
Vo is initial velocity
V is final velocity
a is acceleration
x is distance


The Attempt at a Solution


I'm honestly stumped on this question. I know the objective is to find the max amount possible for d(or x) to be for the man to get to his car safely.
 

Answers and Replies

  • #2
PeterO
Homework Helper
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1. The problem statement, variables and given/known data
A tourist being chased by an angry bear is running in straight line toward his car at a speed of 4.0m/s. The car is a distance d away. The bear is 26m behind the tourist and running at 6.0m/s. The tourist reaches the car safely. What is the maximum possible value for d?


Homework Equations


V = Vo + at
X = 1/2(Vo + V)t
X = Vo * t + 1/2at^2
V^2 = Vo^2 + 2ax
Vo is initial velocity
V is final velocity
a is acceleration
x is distance


The Attempt at a Solution


I'm honestly stumped on this question. I know the objective is to find the max amount possible for d(or x) to be for the man to get to his car safely.
At what rate is the bear gaining on the tourist?
How far from the tourist was the bear when this chase began?
How long will it take the bear to catch the man?
How far will the man run in that time?
What was at the end of that distance?
 
  • #3
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At what rate is the bear gaining on the tourist?
How far from the tourist was the bear when this chase began?
How long will it take the bear to catch the man?
How far will the man run in that time?
What was at the end of that distance?
All of the information I supplied was all that I was given :[
 
  • #4
CAF123
Gold Member
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All of the information I supplied was all that I was given :[
You do have enough information to solve :)
Write equations for the bear's position as a function of time and the tourist's positions as a function of time. It may help to set up a suitable coordinate system.
 
  • #5
PeterO
Homework Helper
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All of the information I supplied was all that I was given :[
I was not asking for information, I was breaking the question down into a series of simple steps that you should be able to do WITHOUT reference to formulas.

Again:

At what rate is the bear gaining on the tourist? Hint: how fast is each of them running?
 
  • #6
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You do have enough information to solve :)
Write equations for the bear's position as a function of time and the tourist's positions as a function of time. It may help to set up a suitable coordinate system.
Gah I got like 10 of these questions T-T. Also that didn't make such sense to me.
 
  • #7
PeterO
Homework Helper
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Gah I got like 10 of these questions T-T. Also that didn't make such sense to me.
Did you read my post #5?
 
  • #8
CAF123
Gold Member
2,906
88
Gah I got like 10 of these questions T-T. Also that didn't make such sense to me.
So the tourist starts 26m ahead of the bear and is running at 4m/s. Use one of the kinematic relations to obtain the tourist's position as a function of time. Now do the same for the bear. It may help to clarify things if you set up a coordinate system with the origin at the bear's position. So the tourist's initial position (i.e [itex] x_o [/itex]) is 26m, while the bears is 0.
Once you have these expressions equate them to find the time it takes for them to meet. PeterO has outlined the method of calculation.
 
  • #9
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The bear was gaining at a rate of 2.0m/s? Right? If so than what would be my next step?
 
  • #10
PeterO
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The bear was gaining at a rate of 2.0m/s? Right? If so than what would be my next step?
Correct rate of gaining.

Read my 5 questions .. but

How much ground did the bear have to make up to catch the tourist, and at at that rate; how long will that take?

Edit: about to go out, but I emphasise - read my 5 questions in post #2
 
  • #11
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26m to make up while at a gain rate of. 2.0 m/s soo I'm not sure but.....
6.0m/s for 26m is 4.33s
2m/s for 13s
Sort of lost at which time to use.
 
  • #12
PeterO
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26m to make up while at a gain rate of. 2.0 m/s soo I'm not sure but.....
6.0m/s for 26m is 4.33s
2m/s for 13s
Sort of lost at which time to use.
Well that's a pity.

Does the 26 metres represent the distance to travel to catch the tourist, or the extra distance to travel to catch the tourist?

You would thus have to chose whether to use the speed of the bear, or the extra speed of the bear, when calculating the time to make up those 26 m.

Hint: when the bear has run 26m, where is the tourist?
 
  • #13
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0
I would guess that the 26m represents the distance travel to catch the tourist and would have to use the speed of the bear(not the 2.0m/s) so it would take 4.33s for the bear to catch the tourist. Now with the new time variable(t) I guess I can now plug in the variables into an equation.

Formula?
X = 1/2(Vo+V)t

x = .5(4.0*6.9)4.33

??????
 
  • #14
PeterO
Homework Helper
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I would guess that the 26m represents the distance travel to catch the tourist and would have to use the speed of the bear(not the 2.0m/s) so it would take 4.33s for the bear to catch the tourist. Now with the new time variable(t) I guess I can now plug in the variables into an equation.

Formula?
X = 1/2(Vo+V)t

x = .5(4.0*6.9)4.33

??????
You're not guessing very well.

While the bear runs 26m [in 4.433 seconds] what is the tourist doing? Now I know the Tourist in Denali National park was standing there taking pictures, but our tourist is running at 4 m/s. What will he achieve in 4.33 seconds? What will the bear have to do after it has run 26m?
 
  • #15
PeterO
Homework Helper
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I would guess that the 26m represents the distance travel to catch the tourist and would have to use the speed of the bear(not the 2.0m/s) so it would take 4.33s for the bear to catch the tourist. Now with the new time variable(t) I guess I can now plug in the variables into an equation.

Formula?
X = 1/2(Vo+V)t

x = .5(4.0*6.9)4.33

??????
Imagine you are going to replicate this situation on a large table - like the ones they use in war rooms, with a little figure of a man, and a little figure of a bear.

Being to scale, you place the figures 26 cm apart.
Now you move the man 4 cm from the bear, then the bear 6 cm towards the man.
That represents the first second of the chase.
How far apart will the models be?
You repeat those moves - man 4 cm from bear, then bear 6cm towards man.
That represents the second second of the chase.
How far apart will the models be?
How far has the man moved so far?
You repeat the moves - man 4 cm from bear, then bear 6cm towards man.
That represents the third second of the chase.
How far apart will the models be?
How far has the man moved so far?
Do you see a pattern emerging?
If you do - solve the problem.
If you dont: -
You repeat the moves - man 4 cm from bear, then bear 6cm towards man.
That represents the fourth second of the chase.
How far apart will the models be?
How far has the man moved so far?
Do you see a pattern emerging?
If you do - solve the problem.
If you dont: -
You repeat the moves - man 4 cm from bear, then bear 6cm towards man.
That represents the fifth second of the chase.
How far apart will the models be?
How far has the man moved so far?
Do you see a pattern emerging?
If you do - solve the problem.
If you dont: -
You repeat the moves - man 4 cm from bear, then bear 6cm towards man.
That represents the sixth second of the chase.
How far apart will the models be?
How far has the man moved so far?
Do you see a pattern emerging?
If you do - solve the problem.
If you dont: - keep repeating until the bear catches the man, and consider how far the man has moved in total.
 

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