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urgent maths paper 4 help needed

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For A, integrate the first equation v = A (t - 0.05t^2) with respect to t, with boundary t=0 to t=15. Integrating the velocity function gives us the distance, so, equate the integral you just found to 225, as the distance from t=0 to t=15 is 225m.

For B, since v = A (t - 0.05t^2) and v = B/t^2 both when t=15, just equate these two equations using the value of A found as above and t=15.
 
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For A, integrate the first equation v = A (t - 0.05t^2) with respect to t, with boundary t=0 to t=15. Integrating the velocity function gives us the distance, so, equate the integral you just found to 225, as the distance from t=0 to t=15 is 225m.

For B, since v = A (t - 0.05t^2) and v = B/t^2 both when t=15, just equate these two equations using the value of A found as above and t=15.

is it the answer for maths 0580/41/m/j/2010 question
 
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Hm, so I see that 0580/m/j/10: There is no such part as Q7(a)(i).. there is only Q7(a), to which I answer:

First convert the radius of 31cm to m, that is 0.31m. Volume of cylinder is pi*r^2*h = 3.142*0.31*0.31*15 = 4.53 m^3.
 
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