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2009-01-18 14:21:46
TheunforgivenII
Member
+12|6703
I have two problems that I cannot solve after trying and trying and been unsuccessful, so here are the two problems and hope some people can help.

First one:

1) Starting from home, you bicycle 21 km north in 2.2 h, then turn around and pedal straight home in 1.1 h. 

(a) What is your displacement at the end of the first 2.2 h?
      my answer: 21 km (North)
(b) What is your average velocity over the first 2.2 h?
      my answer: 10 Km/h (North)
(c) What is your average velocity for the homeward leg of the trip?
      my answer: 19 km/h (South)
(d) What is your displacement for the entire trip?
      my answer: 29 km (North)
(e) What is your average velocity for the entire trip?
      my answer: 9 km/h (North)

b, d and e are wrong for sure because I'm using WebAssign and it said so.

Second one:

1) A gazelle accelerates from rest at 3.7 m/s^2 over a distance of 60 m to outrun a predator. What is its final speed?

I have entered an answer yet on this because I've been working on it on paper and kept on getting different numbers. And like a stupid person like me whenever I get the wrong number on scratch paper I throw it away and now have no clue what to do. When I'm looking at these problems they look easy but I can't get them...I might be over thinking it, but I need help if anybody can do them and thanks in advance
2009-01-18 14:23:41
Bevo
Nah
+718|6942|Austin, Texas
d) is 0
e) 9km/h S
2009-01-18 14:26:13
liquidat0r
wtf.
+2,223|7049|UK
b is 9.545454, surely

d) Displacement is 0.
2009-01-18 14:28:37
TheunforgivenII
Member
+12|6703

MadKatter wrote:

d) is 0
e) 9km/h S
Awesome (d) is correct but why is it 0 though? I do not quite understand what displacement means. Oh and answer to (e) has to be north
2009-01-18 14:29:57
Surgeons
U shud proabbly f off u fat prik
+3,097|6911|Gogledd Cymru

2.2 hours, by that do you mean 2 hours 20 minutes or 2 hours 12 minutes?

b) when using 2 hours 20 minutes = 2.5 m/s (9 Km/h)
    when using 2 hours 12 minutes = 2.65 m/s (9.55 Km/h)

c) when using 2 hours 20 minuted = 5 m/s (18 Km/h)
    when using 2 hours 12 minutes = 5.3 m/s ( 19.1 Km/h)

d) 0, where did you get 29 from?

e) when using 3 hours 30 minutes = 3.33 m/s ( 12 Km/h)
    when using 3 hours 18 minutes = 3.54 m/s ( 12.73 Km/h)

Edit: It's been awhile since I did physics :p

Last edited by The Sheriff (2009-01-18 14:30:53)

2009-01-18 14:30:22
Bevo
Nah
+718|6942|Austin, Texas

TheunforgivenII wrote:

MadKatter wrote:

d) is 0
e) 9km/h S
Awesome (d) is correct but why is it 0 though? I do not quite understand what displacement means. Oh and answer to (e) has to be north
Displacement is distance from the starting position.

E) is -9km/h N then.
2009-01-18 14:30:26
liquidat0r
wtf.
+2,223|7049|UK
Displacement is how far away you are from where you started. Nothing to do with distance travelled.
2009-01-18 14:31:40
Bevo
Nah
+718|6942|Austin, Texas

The Sheriff wrote:

2.2 hours, by that do you mean 2 hours 20 minutes or 2 hours 12 minutes?

b) when using 2 hours 20 minutes = 2.5 m/s (9 Km/h)
    when using 2 hours 12 minutes = 2.65 m/s (9.55 Km/h)

c) when using 2 hours 20 minuted = 5 m/s (18 Km/h)
    when using 2 hours 12 minutes = 5.3 m/s ( 19.1 Km/h)

d) 0, where did you get 29 from?

e) when using 3 hours 30 minutes = 3.33 m/s ( 12 Km/h)
    when using 3 hours 18 minutes = 3.54 m/s ( 12.73 Km/h)

Edit: It's been awhile since I did physics :p
Congratulations on getting every question wrong

ed: you ninja'd D.

Last edited by MadKatter (2009-01-18 14:34:43)

2009-01-18 14:32:31
Jenspm
penis
+1,716|7154|St. Andrews / Oslo

TheunforgivenII wrote:

MadKatter wrote:

d) is 0
e) 9km/h S
Awesome (d) is correct but why is it 0 though? I do not quite understand what displacement means. Oh and answer to (e) has to be north
you haven't moved anywhere, therefore your displacement (change of position) is 0 (you went somewhere and back)
https://static.bf2s.com/files/user/26774/flickricon.png https://twitter.com/phoenix/favicon.ico
2009-01-18 14:34:43
TheunforgivenII
Member
+12|6703

The Sheriff wrote:

2.2 hours, by that do you mean 2 hours 20 minutes or 2 hours 12 minutes?

b) when using 2 hours 20 minutes = 2.5 m/s (9 Km/h)
    when using 2 hours 12 minutes = 2.65 m/s (9.55 Km/h)

c) when using 2 hours 20 minuted = 5 m/s (18 Km/h)
    when using 2 hours 12 minutes = 5.3 m/s ( 19.1 Km/h)

d) 0, where did you get 29 from?

e) when using 3 hours 30 minutes = 3.33 m/s ( 12 Km/h)
    when using 3 hours 18 minutes = 3.54 m/s ( 12.73 Km/h)

Edit: It's been awhile since I did physics :p
well on (d) I guessed because I didn't know what displacement meant because I'm stupid and (c) is already correct.
2009-01-18 14:35:57
liquidat0r
wtf.
+2,223|7049|UK
Just realised I completely misread your second question. (Gazelle)

s = 60 m
u = 0 ms-1
v = ? ms-1
a = 3.7 ms-2
t = Unknown, so don't use

Where:

s = Displacement
u = Initial velocity
v = Final velocity
a = acceleration
t = time

Use:

v2 = u2 + 2as
v2 = 0 + 2as
v = sqrt(2as)

And you get: v = 21.07 ms-1
2009-01-18 14:37:05
argo4
Stand and Deliver
+86|6354|United States
e would be 0 because velocity is displacement(0) over time
2009-01-18 14:39:52
TheunforgivenII
Member
+12|6703

MadKatter wrote:

TheunforgivenII wrote:

MadKatter wrote:

d) is 0
e) 9km/h S
Awesome (d) is correct but why is it 0 though? I do not quite understand what displacement means. Oh and answer to (e) has to be north
Displacement is distance from the starting position.

E) is -9km/h N then.
I just entered -9 km/h but it was wrong, I don't think it should be negative?
2009-01-18 14:40:56
Bevo
Nah
+718|6942|Austin, Texas

argo4 wrote:

e would be 0 because velocity is displacement(0) over time
This, dur-de-dur.

2009-01-18 14:41:55
Gawwad
My way or Haddaway!
+212|7106|Espoo, Finland

TheunforgivenII wrote:

rong for sure because I'm using WebAssign and it said so.

Second one:

1) A gazelle accelerates from rest at 3.7 m/s^2 over a distance of 60 m to outrun a predator. What is its final speed?
I'll give you a hint:
S=S0 + V0t + 1/2At2
Where
S= Distance
S0= Distance in the beginning (for your problem you can ignore this)
V0= Starting velocity
A= Acceleration
t= Time

Never really studied physics, hope that's even correct

Last edited by Gawwad (2009-01-18 14:45:54)

2009-01-18 14:46:40
TheunforgivenII
Member
+12|6703

liquidat0r wrote:

Just realised I completely misread your second question. (Gazelle)

s = 60 m
u = 0 ms-1
v = ? ms-1
a = 3.7 ms-2
t = Unknown, so don't use

Where:

s = Displacement
u = Initial velocity
v = Final velocity
a = acceleration
t = time

Use:

v2 = u2 + 2as
v2 = 0 + 2as
v = sqrt(2as)

And you get: v = 21.07 ms-1
thanks again...though which equation is that? The acceleration or displacement equation.
2009-01-18 14:49:12
liquidat0r
wtf.
+2,223|7049|UK

TheunforgivenII wrote:

thanks again...though which equation is that? The acceleration or displacement equation.
http://en.wikipedia.org/wiki/Equations_ … Derivation
2009-01-18 14:54:05
TheunforgivenII
Member
+12|6703

liquidat0r wrote:

TheunforgivenII wrote:

thanks again...though which equation is that? The acceleration or displacement equation.
http://en.wikipedia.org/wiki/Equations_ … Derivation
oh they are Derivatives....haven't learned them yet, which makes me think that I should of taken Calculus first then Physics
2009-01-18 14:56:35
liquidat0r
wtf.
+2,223|7049|UK
No, derivation just means they're taking some, erm, standard equations that just happen to exist and rearranging them to get rid of some unknown variables.
2009-01-18 14:56:48
Vub
The Power of Two
+188|6916|Sydney, Australia
d) is 0 because displacement is the distance between your starting and ending points, which is the same point.

e) is also zero because velocity = displacement/time = 0/time = 0
2009-01-18 15:07:05
TheunforgivenII
Member
+12|6703

liquidat0r wrote:

No, derivation just means they're taking some, erm, standard equations that just happen to exist and rearranging them to get rid of some unknown variables.
is that the same as in Calculus?
2009-01-18 15:10:44
liquidat0r
wtf.
+2,223|7049|UK

TheunforgivenII wrote:

liquidat0r wrote:

No, derivation just means they're taking some, erm, standard equations that just happen to exist and rearranging them to get rid of some unknown variables.
is that the same as in Calculus?
Based on the problems you just asked about, I'd say that calculus is beyond the scope of your course and you just need to learn those basic equations of motions and be able to rearrange them and substitute in for unknown variables.

Sure, they come from calculus, but I would have thought that you just need to know that they exist and not worry about where they come from.
2009-01-18 16:03:52
Scorpion0x17
can detect anyone's visible post count...
+691|7187|Cambridge (UK)
Are we talking Newtonian or Einsteinian physics?

'Cos if it's the later, all your answers are wrong.

But I'm guessing it's the former.
2009-01-18 20:04:34
TheunforgivenII
Member
+12|6703

liquidat0r wrote:

TheunforgivenII wrote:

liquidat0r wrote:

No, derivation just means they're taking some, erm, standard equations that just happen to exist and rearranging them to get rid of some unknown variables.
is that the same as in Calculus?
Based on the problems you just asked about, I'd say that calculus is beyond the scope of your course and you just need to learn those basic equations of motions and be able to rearrange them and substitute in for unknown variables.

Sure, they come from calculus, but I would have thought that you just need to know that they exist and not worry about where they come from.
i see, i think i understand and thanks for the equation
2009-01-18 20:08:49
Defiance
Member
+438|7092

Scorpion0x17 wrote:

Are we talking Newtonian or Einsteinian physics?

'Cos if it's the later, all your answers are wrong.

But I'm guessing it's the former.
This is all Physics 101, did you really need to ask or are you just looking to argue with someone?

Happens too often around here.

 

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