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[Zrk2]

I just fucked with proportionality in physics. That shit is crazy.

[Simmura McCrea]

Two questions I'm needing help on, both applied maths/physics.

9. A particle starts with a velocity V and moves in a straight line with constant acceleration a. When the velocity reaches 5V, the particle is subjected to a constant retardation a. Show that when the particle returns to its starting position its velocity is -7V.

10. The times taken by a particle to travel two successive distances of x metres are t₁ and t₂ seconds respectively. Show that the acceleration, assumed constant, is given by (2x(t₁-t₂))/(t₁t₂(t₁+t₂))

For 9, I think the displacement is 12V²/a and the time taken for that is 4V/a but where I go from that is a bit of a mystery.

[Urist Imiknorris]

Well, for 9, if a is constant, then v(t) = at + v₀ and x(t) = at²/2 + v₀t + x₀, with v₀ = V in both cases and x₀ = 0.

So set v(t) = 5V and you get t=4V/a.
Plug that into the distance formula and you'll get x(4V/a) = (4V/a)²a/2 + 4V²/a = 12V²/a.
So now you have the distance it traveled with acceleration a, so plug that back into the distance equation as x₀ and 5V as v₀, then set x = 0 and solve for the time it takes to get back to the starting point. Plug your new t and v₀ into the velocity equation to find your final velocity at that time/point.

[Simmura McCrea]

x(t) = at²/2 + v₀t + x₀

That makes some sense, but the quoted formula doesn't. It resembles the s = ut + 1/2at² formula, but the x₀ bit is completely unfamiliar. Could you explain that?

[Urist Imiknorris]

How much calculus have you taken?

[Simmura McCrea]

UK advanced higher maths level only. I'm not too sure what the US equivalent is, but it's 6th year of high school maths over here.

[ed boy]

Two questions I'm needing help on, both applied maths/physics.

9. A particle starts with a velocity V and moves in a straight line with constant acceleration a. When the velocity reaches 5V, the particle is subjected to a constant retardation a. Show that when the particle returns to its starting position its velocity is -7V.

10. The times taken by a particle to travel two successive distances of x metres are t₁ and t₂ seconds respectively. Show that the acceleration, assumed constant, is given by (2x(t₁-t₂))/(t₁t₂(t₁+t₂))

For 9, I think the displacement is 12V²/a and the time taken for that is 4V/a but where I go from that is a bit of a mystery.

9: Find the distance from the start when the particle first reaches 5V. From then, construct another quadratic equation, with initial position your previous answer, initial velocity 5V, and acceleration -a. Find how long it takes to reach the starting point (when position=0) and use that time to find the speed.

10:Let V1 be the velocity at the start of the first stretch, V2 be the velocity between them and V3 be the velocity and the end of the second. We know that acceleraion is constant, and that s=ut+at²/2, so we can see that:

2x=2V1T1+AT1²
2x=2V2T2+AT2²
2V1T1+AT1²=2V2T2+AT2²
A(T1²-T2²)=2V2T2-2V1T1
A=2V2T2/(T1²-T2²) (Assuming that T1 is not equal to T2. If it is, then acceleration is zero).

[Simmura McCrea]

Can someone please enlighten me on this formula involving initial position and why I've never been taught it in maths or physics?

[MonkeyHead]

Teaching this stuff means I can comment with some authority. At AS and A2 Level for some reason the equations of motion are only taught with reference to "displacement" - that is, a raw change in s (or x, depending on notation), not in relation to any other point or body. The initial position term is more useful when considering an objects position relative to an external reference frame - some fixed point or a second object. It helps to consider that there is no such thing as absolute motion, so when giving an objects position (or change in position) you should really be giving its position in relation to something else.

[Simmura McCrea]

So in order to answer the question, I have to use stuff I've never been taught and that isn't in the stuff given for the course? Frigging frig. I'm gonna drop this on my teacher, see what he says. In the meantime, there's another question.

A stone of mass 0.4kg is thrown vertically into the air with an initial velocity of 20ms^-1. If there is a "drag" force resisting the motion of 0.2N, find how far the stone travels before it comes to rest. Also calculate the overall time taken for it to return to its point of release.

[Urist Imiknorris]

So in order to answer the question, I have to use stuff I've never been taught and that isn't in the stuff given for the course?

Nope. I was just taught to use the absolute version.

[Leafsnail]

You don't really need it though, surely.  Just find how far it moves relative to the initial position then add that to the initial position.

The new question is just gonna be SUVATs, I think.  You'd just need to have two separate equations for going up and going down (so on the way up you'd set the initial velocity as 20, the final velocity as 0, the acceleration as the appropriate gravitational force + the drag and so on).

[Reiina]

I was wondering if I could still solve a high school problem(the 9) problem) and I can. Hooray for me :p.

Spoiler (click to show/hide)

First equation:
x(t)=(1/2)*at²+Vt
v(t)=at+V

t1 is when v(t1)=5V

v(t1)=5V
a(t1)+V=5V
(t1)=4V/a

x(t1)=(1/2)*a*(4V/a)²+V(4V/a)
x(t1)=(1/2)*(16V²/a)+4V²/a)
x(t1)=12V²/a

---
equation starting at time t1
x(t)=-(1/2)*at²+5Vt+x(t1)
x(t)=-(1/2)*at²+5Vt+12V²/a
v(t)=-at+5V

t2 is when x(t2)=0
-(1/2)*at²+5Vt+12V²/a=0
delta=(5V)²-4*-1/2*a*12V²/a=25V²+2*12V²=49V²=(7V)²
t2=(-5V-7V)/-a=12V/a
t2=(-5V+7V)/-a=-2V/a
since t2>0 t2=12V/a

v(t2)=-12V+5V=-7V

[Simmura McCrea]

I was wondering if I could still solve a high school problem(the 9) problem) and I can. Hooray for me :p.

Spoiler (click to show/hide)

First equation:
x(t)=(1/2)*at²+Vt
v(t)=at+V

t1 is when v(t1)=5V

v(t1)=5V
a(t1)+V=5V
(t1)=4V/a

x(t1)=(1/2)*a*(4V/a)²+V(4V/a)
x(t1)=(1/2)*(16V²/a)+4V²/a)
x(t1)=12V²/a

---
equation starting at time t1
x(t)=-(1/2)*at²+5Vt+x(t1)
x(t)=-(1/2)*at²+5Vt+12V²/a
v(t)=-at+5V

t2 is when x(t2)=0
-(1/2)*at²+5V+12V²/a=0
delta=(5V)²-4*-1/2*a*12V²/a=25V²+2*12V²=49V²=(7V)²
t2=(-5V-7V)/-a=12V/a
t2=(-5V+7V)/-a=-2V/a
since t2>0 t2=12V/a

v(t2)=-12V+5V=-7V

I follow you right up until those last six lines then completely lose you. None of those 6 lines make any sense to me. It's probably because I'm tired. Could you (or anyone, really) explain them?

Also, it's university work. Albeit,  1st year revisionary stuff.

[Reiina]

I follow you right up until those last six lines then completely lose you. None of those 6 lines make any sense to me. It's probably because I'm tired. Could you (or anyone, really) explain them?

Also, it's university work. Albeit,  1st year revisionary stuff.

Oops just noticed I forgot a t in there

well the last lines are simply solving -(1/2)*at²+5Vt+12V²/a=0 which is a 2nd degree equation(ax²+bx+c=0).
With the solutions being (-b+sqrt(b²-4ac))/2a and (-b-sqrt(b²-4ac))/2a.
What I call delta is b²-4ac btw.