hi, can someone explain me the Q4. ii) of this paper. i know how to calculate it, but my answer gives 50ms^-1, while the march scheme shows another way and gives 47ms^-1. could someone do it , and explain wat answer they got?? 50ms^-1 like me or 47.5ms^-1 as march schme. thanks
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Mechanics paper 5 O/N 2007
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okay after you have proved that Acceleration = (10-0.2V)
A = dv/dt
dv/dt = 10-0.2v
dv=0.2(50-v)dt
5/(50-v) dv = dt
Integrate with limits 0 to v for the dv and 0 to 15 for dt
You should have this result:
5[-ln|50-v|] from 0 to v = t from 0 to 15
5[-ln|50-v|- -ln|50|]=15
-ln|50-v| + ln|50| = 3
-ln |50-v| = 3-ln|50| = -0.91202...
By taking e for both sides
|50-v|^-1 = 0.40171..
|50-v| = 1/Ans = 2.4893...
v=47.5 m/s
When i first tried to solve it, i forgot to divide the ln|50-v| by -1,
which is the diff or the bracket
Good Luck
!
A = dv/dt
dv/dt = 10-0.2v
dv=0.2(50-v)dt
5/(50-v) dv = dt
Integrate with limits 0 to v for the dv and 0 to 15 for dt
You should have this result:
5[-ln|50-v|] from 0 to v = t from 0 to 15
5[-ln|50-v|- -ln|50|]=15
-ln|50-v| + ln|50| = 3
-ln |50-v| = 3-ln|50| = -0.91202...
By taking e for both sides
|50-v|^-1 = 0.40171..
|50-v| = 1/Ans = 2.4893...
v=47.5 m/s
When i first tried to solve it, i forgot to divide the ln|50-v| by -1,
which is the diff or the bracket
Good Luck
!