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# Help with M1 integration watch

1. To put it simply... I don't get it.

I'm doing the OCR course and my exam is a week today, but I still can't get my head around integration...

Looking at my notes, I have figured that:

a (acceleration) integrates into v
v (velocity) integrates into s

I don't understand WHY this happens and I've searched the whole web and there is no AS site with anything about integration in the M1 area.

Here's an example question... I have the answers so I'm not interested in them I want to know HOW you worked them out and WHY the logic you apply to them works.

QUESTION:
A particle P moves in a straight line so that, at time t seconds afer leaving a fixed point O, it's acceleration is -1/10t ms/\-2. At time t=0, the velocity of P is Vms/\-1.

(i) Find, by integration, an expression in terms of t and V for the velocity P.
(ii) Find the value of V, given that P is instantaneously at rest when t=10.
(iii) Find the displacement of P from 0 when t=10.

2. (Original post by Eye Jay)
To put it simply... I don't get it.

I'm doing the OCR course and my exam is a week today, but I still can't get my head around integration...

Looking at my notes, I have figured that:

a (acceleration) integrates into v
v (velocity) integrates into s

I don't understand WHY this happens and I've searched the whole web and there is no AS site with anything about integration in the M1 area.

Here's an example question... I have the answers so I'm not interested in them I want to know HOW you worked them out and WHY the logic you apply to them works.

QUESTION:
A particle P moves in a straight line so that, at time t seconds afer leaving a fixed point O, it's acceleration is -1/10t ms/\-2. At time t=0, the velocity of P is Vms/\-1.

(i) Find, by integration, an expression in terms of t and V for the velocity P.
(ii) Find the value of V, given that P is instantaneously at rest when t=10.
(iii) Find the displacement of P from 0 when t=10.

I'm not sure why it happens, but someone will probably tell you....as for the question:

I'm not too sure what the acceleration is....I'm taking it as (-1/10)t rather than this -1/(10t) - i hope it's right

(i) v = INT a.dt = -1/10 [ INT t dt]

v = (-t^2/20) + c

t=0, v=V
so V= 0 + c, c=V

v = V - (t^2/20)

(ii) t=10, v=0
so, 0 = V - (100/20)
0 = V - 5
so V = 5 m/s

(iii) you can now re-write v as: v = 5 - (t^2/20)
x = INT v.dt

x = INT[5 - (t^2/20)]dt

x = 5t - (t^3/60) + k

t=0, x=0, so k=0

x = 5t - (t^3/60)

t=10, x = 50 - (1000/60)
x = (100/3) m
3. (Original post by Eye Jay)
To put it simply... I don't get it.

I'm doing the OCR course and my exam is a week today, but I still can't get my head around integration...

Looking at my notes, I have figured that:

a (acceleration) integrates into v
v (velocity) integrates into s

I don't understand WHY this happens and I've searched the whole web and there is no AS site with anything about integration in the M1 area.
Velocity is defined as the rate of change of displacement, so v = ds/dt (think about speed = distance/time, it's the same here).
Acceleration is defined as the rate of change of velocity, so a = dv/dt (think about acceleration = change in velocity/time). Obviously integrating both those expressions wrt time will give you displacement and velocity.

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Updated: June 7, 2004
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