A as a function of X (1 Viewer)

[CrashOveride]

acceleration as a function of position.

A particle is moving in a straight line with constant accel. of 2m.s^-2. If it starts from the origin with a velocity of -6m/s, find

a) An expression for the velocity in terms of the displacement.

Here's my working:

v² = 2 INT a dx
= 2(2x) + C

v² = 4x + C
36 = 0 + C
C = 36

v² = 4x + 36
v² = 4(x + 9)
v = +/- 2(x+9)^1/2

The condition v = -6 when x = 0 is satisfied by the negative solution.

Back of book says positive solution.

 

[CM_Tutor]

Back of book is wrong. The answer v = 2 * sqrt(x + 9) cannot give v = -6 at x = 0, thus it cannot be the correct solution.

 

[George W. Bush]

is this from Fitzpatrick? The Fitzpatrick projectile motion section has a lot of errors regarding the +/-

 

[CrashOveride]

No it's not from Fitz - it's from Jones & Couchman.

Ok so the negative solution is the one which satisfies the criteria when x=0, v=-6. However, if we think about the behaviour of this particle, noticing it has constant positive acceleration, it will soon slow down and come to rest before changing direction, crossing the origin and then remaining positive velocity when t appraoched infinity.

This is obviously not satisfied by a solely negative solution of v then.

 

[CrashOveride]

Hmmmmm

Hmm is the question fundamentally wrong? Or have i missed something :mad1:

 

[Xayma]

Originally posted by CrashOveride
Hmm is the question fundamentally wrong? Or have i missed something :mad1:

For each non-positive x value it will have two values (exlcuding the end one), it is satisified by the positive negative when x<=0 however when x>0 it is only satisfied by the positive answer.

 

[CrashOveride]

Agreed

 

[CM_Tutor]

Originally posted by CrashOveride
Ok so the negative solution is the one which satisfies the criteria when x=0, v=-6. However, if we think about the behaviour of this particle, noticing it has constant positive acceleration, it will soon slow down and come to rest before changing direction, crossing the origin and then remaining positive velocity when t appraoched infinity.

This is obviously not satisfied by a solely negative solution of v then.

Good point. So, the answer is v = - 2 * sqrt(x + 9) for t < 3, and v = 2 * sqrt(x + 9) for t => 3

 

[CrashOveride]

I found that with that particular exercise, this issue came up quite a bit and the back of the book only addressed one aspect of it..i imagine just by looking at the given data and end of story.

i thought this would be straight mechanical type work, seems like with any of these type of questions u have to keep in mind what the actual particle is doing, in order to devise a logical approach

 

[CrashOveride]

Originally posted by CM_Tutor
Good point. So, the answer is v = - 2 * sqrt(x + 9) for t < 3, and v = 2 * sqrt(x + 9) for t => 3

At t = 3, v = 0. So the sign infront of the expression in v would't really count here? Or do you include it in the positive solution because its non-negative..

 

[CM_Tutor]

My choice of which expression to take for t = 3 was arbitrary - I chose the positive case for aesthetic reasons.