kinematic graph
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hi all, i am stuck at question (c), to find acceleration for 1st 2 hours from the distance time graph.
do look at the link to the question
http://img32.imageshack.us/i/15165090.jpg/
http://img177.imageshack.us/i/73263681.jpg/ the time on the x axis is 0900,1000, 1100, 1200 & 1300
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Could you check your second link? It seems not to be working.
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Originally posted by ThunderFbolt:
Could you check your second link? It seems not to be working.
My physics is half-baked one, so I cannot say that my answer is definitely correct.
1st 2 hours, a = constant ( uniform )
a= (V - U) / t = (10km/h - 0 ) / 2h = 5km/h^2
Where did the 10km/h comes from, you ask? He travelled 20km in 2 hours, so it is 20km/2h = 10km/h.
It will be better for you to wait for the Physics pros to arrive before taking my answer seriously.
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Let me try to answer....not also so sure
Initial speed for the 2 hours.......0 km/h
Final speed for the 2 hours........gradient at 2 hours spot----draw a tangent
Accerlation: Final-initial / time
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question: can the tangent be drawn at any time from 0900 to 1100 hrs. how come only draw tangent at 1100hrs?
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Originally posted by Divanhot:
question: can the tangent be drawn at any time from 0900 to 1100 hrs. how come only draw tangent at 1100hrs?
Because he want the final velocity ( in the time period of interest) aka the velocity when time is 1100hrs.Drawing a tangent is too troublesome and it would only be an estimate.
Forbiddensinner's answer is flawed too. Distance travelled in 1st hr is 5km and 2nd hour is 15km. So it is technically flawed despite value being correct.
If the acceleration continue into the 3rd hr, the distance travelled in that 3rd hour would be 25km. Giving us 45km over 3hrs. Using (45-0)/3, we would get 15km/h then which is wrong.
Perhap you can consider the fact that
velocity = (area under the acceleration time graph)
Since it is uniform acceleration
Area = rectangle
20 = (x)(2)
x = 10
acceleration = 10km^2/h
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hi i will like to use drawing of tangent at 1100 to get the velocity
however, i find it hard to do so as the graph is a curve from 0900-1100 and a straight line from 1100 onwards. hence if i draw a tangent, the tangent will be above the straight line after 1100
hi bro mike, how do you know the acceleration is constant? and how do you obtain 20km/hr for velocity value. thanks
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Originally posted by Divanhot:
hi i will like to use drawing of tangent at 1100 to get the velocity
however, i find it hard to do so as the graph is a curve from 0900-1100 and a straight line from 1100 onwards. hence if i draw a tangent, the tangent will be above the straight line after 1100
hi bro mike, how do you know the acceleration is constant? and how do you obtain 20km/hr for velocity value. thanks
The problem is you can't for 2 reasons.i) The question say calculate not estimate
ii) To draw a tangent, you need to know how the curve behave before and after the point of interest. Therefore you can't draw a tangent to estimate in this case because 1100hrs is the extreme of the domain drawn.
How do I know the acceleration is constant? The question say that acceleration is uniform. uniform = same thru the time period.
As for 20km/h, it is given in the final sentence of the question. "He then contimue... at 20km/h." Indicating that this is the speed at 1100hr.
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hi the way i interpret the question is the constant speed of 20km/hr is only after 1100 hrs( sentence says.. He then continue... at 20km/h), so how can we use velocity= 20km/hr for time between 0900 and 1100hrs?
i accept acceleration is uniform.
can i use s= ut + 1/2 at^2
sub u=o, 20= 1/2 a (2^2)
a= 10m/s^2
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Originally posted by Divanhot:
hi the way i interpret the question is the constant speed of 20km/hr is only after 1100 hrs( sentence says.. He then continue... at 20km/h), so how can we use velocity= 20km/hr for time between 0900 and 1100hrs?
i accept acceleration is uniform.
can i use s= ut + 1/2 at^2
sub u=o, 20= 1/2 a (2^2)
a= 10m/s^2
1st of all, I didn't use velocity = 20km/h between 0900 and 1100... I used velocity = 20km/h at the instant that time = 1100hrs. At that instant the car has sped up to 20km/h and THEN continued at that constant speed.
And yes, you can use s = ut+1/2 at^2 as it is valid. As I understand, alternative approaches using formulas not in syllabus are accepted as long as they are mathematically correct. I was considering showing you this approach but decided not to as they don't teach this in O level A/E maths anymore. And not everyone take physics. Personally I didn't. As I recall, the formula that you use applies to constant acceleration situations.
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thanks, mike! though i still dun understand how come we dun consider velocity from 0900-1100 hrs as we wanted to find acceleration for the 1st 2 hrs( i.e. 0900-1100hrs)
