# M2 Mechanics EdExcel Review Exercise 1 Question

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16 years ago
#1
Hi, I am really stuck on this M2 question and I do not get how its possible to work out the answer. If you have the M2 book,its Review Exercise 1 page 85 question 10b.

Here is the question 10b) if u dont have the M2 book,

Key:-
=< means greater than or equal to
<= means less than or equal to

10b) A particle P moves in a straight line in such a way that, at time t seconds, its velocity v m/s is given by:

v=12t-3t^2 , for 0=<t<=5

v= (-375)t^(-2), for t>5

When t=0, P is at the point 0. Calculate the displacement of P from 0 when t=6.

[To start this question integrate v gives displacement x]
0
16 years ago
#2
Int[0,X]1dx = Int[0,5] 12t - 3t^2 dt
=> X = [6t^2 - t^3] between[0,5] = 25

therefore Int[25,Y] dx = Int[5,6] -375t^-2 dt
=> Y - 25 = 375/6 - 375/5 => Y = 25/2
0
16 years ago
#3
v=ds/dt
ds/dt=12t-3t^2
integrating,
s=6t^2 - t^3 + C => C=0 @ t=0
i.e. s=6t^2 - t^3, 0<=t<=5

@t=5,
s=6(25) - (125)=150-125=25
s=25m

also,
ds/dt=-375t^(-2), t>5
integrating,
(oops)
s=375t^(-1) + D

@t=5,
25=375/(5) + D
25=75+D
D=-50
s=375/t - 50, t>5

@t=6,
s=375/(6)-50=62.5-50
s=12.5m
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