23-01-2009, 03:13 PM
alexgoei Wrote:On the speed/time chart on page 249, the distance for "b" is 367 metres. With initial velocity and final velocity of 22.22 m/s and 11.11 m/s respectively and using Equation 2 I should get 493 metres. May I know how the 367 metres was derived?
Look forward to your answer please.
It looks to me as though you have calculated the distance to stop (0 m/s) rather than 11.11m/s. I cannot tell definitely since I don't have access to either the study pack to see what is written for the equation or your paper at this time, but I suspect it because of the logic below
Values that you know
u = 22.22m/s
v = 11.11m/s
a = -0.5m/s/s (it nearly always is in IRSE exams, don't forget the minus as you are braking)
s = what you are trying to calculate
t = not relevant to this discussion
so, selecting the equations from the standard list, we will use
v^2 = u^2 + 2 a s
11.11^2 = 22.22^2 + 2 (-0.5) s
or 123 = 493 - s
s = 493 - 123 = 370 (which is close enough to 367 depending on your rounding!).
If you set v to 0 m/s to calculate the distance to a stand, you will get s= 493 - 0 which is what you have.
All of the basic equations are summarised in the attachment to my post encouraging people to recall their school maths!
Hope this is helpful.
Peter (M)