AS integration Watch

bigmansouf
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Question:

A particle starts from rest and moves in a straight line. Its speed for the first 3 s is proportional to  (6t-t^2) , where t is the time in seconds from the commencement of motion, and thereafter it travels with uniform speed at the rate it had acquired at the end of the 3rd second. Prove that the distance travelled in the first 3 s is two-thirds of the distance travelled in the next 3 s.


I understand the first 3 seconds is proportional to speed thus
t = 3  \propto   (6t-t^2)
thus speed  = k(6t-t^2)
from here i am stuck I dont know what to do
please help
Last edited by bigmansouf; 1 week ago
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mqb2766
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(Original post by bigmansouf)
Question:

A particle starts from rest and moves in a straight line. Its speed for the first 3 s is proportional to  (6t-t^2) , where t is the time in seconds from the commencement of motion, and thereafter it travels with uniform speed at the rate it had acquired at the end of the 3rd second. Prove that the distance travelled in the first 3 s is two-thirds of the distance travelled in the next 3 s.


I understand the first 3 seconds is proportional to speed thus
t = 3  \propto  (6t-t^2)
thus speed  = k(6t-t^2)
from here i am stuck I dont know what to do
please help
Split it into two parts, up to 3 seconds and then after 3 seconds.
The distance is is the integral of velocity and for the second 3 seconds, the velocity is constant and is equal to the velocity with t=3.
Try setting up those terms and posting if you have problems?
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casperyc
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Intuitively, it's too easy...

You are comparing the following
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bigmansouf
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Report Thread starter 5 days ago
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(Original post by casperyc)
Intuitively, it's too easy...

You are comparing the following
Name:  klf_2019-02-12_19-35_oo7712.600.png
Views: 7
Size:  17.3 KB
thank you
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