Qualification > Math

Last question in the mechanics paper!

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hesho21:
Alright the paper was easy , but I have a doubt on the last question
So A was 4 and B was 3375
so the question said something had two motions, one for 15s and one greater than that , and they satisfy these equations:

V=4t-0.05t2   0<=t<=15

v= 3375/t2 t>15

Distance travelled during the first 15s (first motion is 225m)

now if u integrate the second motion how do u find C?

[Spy]:
the integrated expression is -3375/t + c=225
you know t is 15

so its -225+c=225
c=450

so it asked what is the speed where distance is 315
you do -3375/t+455=315
t=25

then 3375/25^2 which is equal to 5.4

:)

vserian:

--- Quote from: Spy on May 20, 2010, 12:49:34 pm ---the integrated expression is -3375/t + c=225
you know t is 15

so its -225+c=225
c=450

so it asked what is the speed where distance is 315
you do -3375/t+455=315
t=25

then 3375/25^2 which is equal to 5.4

:)

--- End quote ---

But it already travelled 225 m with the first expression for the last forumla to be vails

it shuls be equal to 315-225 = 90

hesho21:

--- Quote from: Spy on May 20, 2010, 12:49:34 pm ---the integrated expression is -3375/t + c=225
you know t is 15

so its -225+c=225
c=450

so it asked what is the speed where distance is 315
you do -3375/t+455=315
t=25

then 3375/25^2 which is equal to 5.4

:)

--- End quote ---


yeh thanx for that clarification , i did it wrong i just put c as 225 , the distance it travelled in motion A :S anyways thnk u :D

neno:
yup i did the exact same mistake as u spy...my bad anyways sooo i need to ask u guys sumthing is it true tht this year thrs not gna b a curve its gna b graded on the bases of percentages??

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