# Mechanics 3 - Further kinematics

1. Mar 3, 2006

### mr bob

I'm studying M3 as one of the three modules for my further maths alevel, and have just started the first chapter. I'm fine with acceleration as a funtion of time, and integrating to find velocity and displacement, and most of acceleration as a function of displacement, but have come stuck on one of the questions:-

A particle p moves along the positive x axis. Its acceleration is (x + 3)ms^-2 when its displacement from the origin O is x meters. Given that initially, when t = 0, the velocity of P is 3ms^-1 in the direction Ox and x = 0. Obtain:-
a) the speed Vms^-1 of P as a funtction of x
b) x as a function of t

I started by trying (1/2)v^2 = INT(X+3)
therefore getting
v^2 = x^2 + 6x + C
where C = 9
therefore:
v^2 = x^2 + 6x + 9

The book's answer however is different, v = x + 3
I dont know how they got that answer, and as for part b) i'm completely lost as i cant figure out how to bring t into the equation.

Any help would be greatly apprectiated,
Thanks,
Bob

2. Mar 3, 2006

### qtp

how can the velocity = x+3 if the acceleration is x+3 ?

3. Mar 3, 2006

### mr bob

I don't know, that's the books answer. It could be wrong though. I know a few were in M1.

4. Mar 3, 2006

### vaishakh

You can find da/dx.
da/dx*v = da/dx*dx/dt = da/dt.

5. Mar 3, 2006

### vaishakh

Your answer is not differant anyway. v^2 = x^2 + 6x + 9 = (x + 3)^2.
Thus v = x + 3. So why should you say that your answer is wrong according to the book? Use the same concept I have shown dor the second question to solve the first question. It is tricky but solve it properly.

6. Mar 3, 2006

### mr bob

Ah of course. Silly mistake there, just wasnt thinking. Thank you.

7. Mar 27, 2009

### holulumaster

(a) d/dx(0.5v2) = ( x + 3 )
=> v2/2 = x2/2 + 3x + C
=> v2 = x2 + 6x + C

when t=0, x=0 and v=3
Therefore, c = 9

v2 = x2 + 6x + 9
=> v2 = (x+3)2
=> v = (x+3)

(b) v = x+3
a = x+3

a = dv/dx * dx/dt
=> (x+3) = 1 * (dx/dt)
Integrating both sides.... ( replace "{" as integration sign... (im a noob in this forum))

=> {dx/x+3 = { dt
=> ln|x+3| = t + c .eqn 1

when t=0, x=0

=> c = ln3

therefore... from eqn 1...

=> ln|x+3| = t + ln3
=> ln|x+3| - ln3 = t
=> ln|(x+3)/3| = t
=> (x+3)/3 = expt
=> x + 3 = 3expt
=> x = 3expt - 3

Last edited: Mar 27, 2009