okay, so i was wondering if anyone here knows how to do physics dimensional analysis. i dont understand it at all and i have homework due on it in a few hours

the question is this using this equation

Find the powers p and q that make this equation dimensionally consistent.

can anyone explain this to me?

Absolutely. So, first, what are you units for v, t, and a?

do you mean [T] [L] and [M]? or do you mean what they actually come out as?

The equation is talking about v=velocity, t=time, and a=acceleration, correct?

yes

So, if we look at the following equation:



So, the units of v are m/s, the units of t are s, and the units of a m/(s^2). So, you need (m/s) == (s)^p * (m/(s^2))^q. So far so good?

okay so far so good

So, all you need is to simplify that out:

(m/s) = ((m^q)*(s^p))/(s^2q)

Because, of course, (a^b)^c = a^bc. We know that we want to end up with one m in the numerator and one s in the denominator. We only have one source of m's, so we know m^q = m, so q=1. This leaves us with (s^p)/(s^2(1)) to give us (1/s). That of course means that 2(1) -1 = p, so p = 1. If we plug these back in to double-check:

v = (t)^1 * (a)^1
v = (s)^1 * (m/(s^2))^1
v = (m*s)/(s*s) = m/s
v = m/s

Success.

ah ha! thanks Kerm, this makes more sense now. i just got one more question. if you have this equation how would you solve this? the "X" dosent seem to pair up with anything. (your looking for P

x is x-position, in meters. x and y are each position, v is velocity, and a is acceleration. t is of course time.

In this case: m/s = m/(s^v). Notice that coefficients can be discarded in dimensional analysis. This is a trivial equation to solve, v=1.

okay that answered my questions! Thanks so much!!!

The pleasure is entirely mine, glad to help.

well, it seems as though i was programming through my physics lesson, so, i kinda need more help.

anyways, hers the problem, well actually theres a lot of problems i need help on, but i post one at a time, maybe i can figure out the rest.

so here it is

A bicyclist is finishing his repair of a flat tire when a friend rides by at 3.6 m/s. Two seconds later, the bicyclist hops on his bike and accelerates at 2.8 m/s2 until he catches his friend.

a) how much time until he reaches his friend?
b)how far has he traveled in this time?
c) what is his speed when he catches up?

Aight, so let's call the guy repairing his tire b, and his friend will be f. Now, at t=0, his friend rides by at 3.6m/s. Therefore, the position of his friend xf = 0 + 3.6t, assuming that the position where b was fixing his bike x = 0. Now, the other guy starts accelerating at ab = 2.8 m/s^2. Now, from the standard kinematic equation with acceleration:

xb = x0 + v0b*t + 0.5*ab*t^2

However, x0=0, v0b=0, and we'll replace t with tb = t-2

xb = 0.5*ab*(t-2)^2

We want to know when xb = xf:

3.6t = 0.5*(2.8)*(t-2)^2
3.6t = 1.4*(t-2)^2
3.6t = 1.4*(t^2-4t+4)
3.6t = 1.4t^2-5.6t+5.6
0 = 1.4t^2-(5.6+3.6)t+5.6
0 = 1.4t^2-9.2t+5.6

Now we apply the quadratic equation:

t = [9.2 +/- sqrt(9.2^2-4(1.4)(5.6)]/2(1.4)
t1 = 5.893
t2 = 0.679

The guy who was fixing his bike didn't start riding until t=2, so it has to be t1 = 5.893. We'll double-check:

xf = 0 + 3.6t = 3.6(5.893) = 21.215
xb = 0.5*ab*(t-2)^2 = 1.4(t-2)^2 = 1.4(3.893)^2 = 21.218

Let's call that close enough! So we know it took him 5.893 seconds from when his friend walked by or 3.893 seconds after he started riding, and he traveled 21.21 meters in that time.

For the final part of the problem:

vb = v0 + ab*tb
vb = 0 + 2.8*(t-2)
vb = 2.8(3.893)
vb = 10.90 m/s

He passes his friend going 10.9 m/s (which is pretty insanely fast btw)

okay, heres two more for you. i hope these are the last...

A dog runs back and forth between its two owners, who are walking toward one another (Figure 2-25). The dog starts running when the owners are d = 9.0 m apart. If the dog runs with a speed of vd = 2.6 m/s, and the owners each walk with a speed of 1.3 m/s, how far has the dog traveled when the owners meet?



and

A rocket blasts off and moves straight upward from the launch pad with constant acceleration. After 2.8 s the rocket is at a height of 80.0 m.
(a) What is the acceleration of the rocket?
(b) What is its speed at this time?

thank you so much Kerm.

Well, let me not do them all for you. See where you can get with those first and show me where you get stuck.

okay good idea

1) Since the dog is going at 2.6m/s, and the owner is walking at 1.3m/s, then they're going towards each other at 3.9m/s. Since they started 9m apart, and they come together at 3.9m/s, then they take 9/3.9 = 2.31 seconds to come together. Since the dog was going 2.6m/s, 2.6*2.31 = 6 meters (which makes sense that the dog covered 2/3 of the distance since it was going twice as fast.

2) The rocket:
2a) t=2.8s, y=80m. Usual kinematics equation: y = y0 + v0t + 0.5at^2
80 = 0 + 0t + 0.5a(2.8^2)
80 = 3.92a
a = 20.4 m/s^2

2b) Then, v = at = (20.4)2.8 = 57.12 m/s.

okay, the answer for the dog problem turned out to be 9 instead of 6. heres what i did

each person is walking half the distance of 9, so 4.5m, the accel. is 0, and speed is 1.3m/s
so I used the equation x=x0+v0t+.5at^2
a=0 and x0=0 so that eliminates a whole part of the equation, so we are left with

x=v0t or
4.5=1.3t

so it takes 3.46 sec for the owners to meet. if the dog was traveling at 2.6 m/s then we take the time * the speed (3.46*2.6) and we get 9!

im finally getting this stuff

Yeah, that was me not reading carefully enough. I thought it was asking when the dog and the right person meet, not the two people. Sorry about that.