Motion on Incline Problem - Find the Distances Marked Every 2.5s

[Cole07]

problem: A ball is allowed to roll from rest down an inclined plane, and the distances are marked every 2.5s. if the second mark is made 1.5m from the starting point, where is the first mark?

where is the fourth mark?

well i made a velocity vs. t chart and the velocity of it is 0.6 then ifound the velocity of ech number on my graph but iwas told my answer was wrong. can you help me please.

 

[Cyrus]

Can you put your homework questions in the homework section please.

 

[Cole07]

how this is only my second time?

 

[Cyrus]

There is a thread for homework help on the main page. It's the second one down from the top.

 

[arildno]

Let indices 1,2 indicate different instants and the associated distances travelled. Thus, using your distance formula for two different times, we've got the formulae:
$$d_{1}=\frac{a}{2}t_{1}^{2} (1), d_{2}=\frac{a}{2}t_{2}^{2}(2)$$
where a is the acceleration.

Now, divide these equations with each other:
$$\frac{d_{1}}{d_{2}}=\frac{\frac{a}{2}t_{1}^{2}}{\frac{a}{2}t_{2}^{2}}\to\frac{d_{1}}{d_{2}}=\frac{t_{1}^{2}}{t_{2}^{2}}\to$$
$$\frac{d_{1}}{d_{2}}=(\frac{t_{1}}{t_{2}})^{2}$$

agreed?

 

[Cole07]

i'm sorry i don't understand any of that

 

[arildno]

Nonsense.
Start studying it from line 1, and report back precisely what you don't understand!

 

[Cole07]

where do you get 1,2

 

[arildno]

I give two NAMES to two different instants!
Am I not allowed to give names to quantities?

Should I give them the same name, perhaps??

 

[Cole07]

so the 1st equation would be 1.5=a(2.5)^2 right?

 

[arildno]

If you want $t_{1}$ to be interpreted as 2.5, yes.

 

[Cole07]

ok for this i get 0.24, but what is the second instants the fourth mark?

 

[arildno]

Look, do you understand how I got the equation:
$$\frac{d_{1}}{d_{2}}=(\frac{t_{1}}{t_{2}})^{2}$$

 

[Cole07]

not really, i do understand the 1,2 now but i don't see where you get the second instants. this problem just really confuses me I'm sorry.

 

[arildno]

Do you understand that I can divide one equation with the other?

 

[Cole07]

so you took d=a(t)^2 and divided it by itself? with different subscripts?

 

[Cole07]

so you took d=a(t)^2 and divided it by itself to get D(1)/D(2)=(t(1)/t(2))^2 right?

 

[Cole07]

sorry i didn't see it come up i typed it again ignore the first on please.

 

[arildno]

so you took d=a(t)^2 and divided it by itself to get D(1)/D(2)=(t(1)/t(2))^2 right?

Correct!
Do you agree I am allowd to do that?

 

[Cole07]

yes, but is the first instants the first mark and the second instants the second mark?

 

[arildno]

Well, they can be!
But, and this is important:
Do you agree that this equation would hold whatever particular instant we let t1 stand for and whichever other instant t2 stand for?

 

[Cole07]

Yes I agree so would d(1)=1.5 meters and t(1) = 2.5 seconds and i still have 2 unknowns I am so sorry to be so dumb but this is my sixth week of physics and I'm just not getting this stuff

 

[arildno]

Well, but that is the beauty of the equation!

To show you how to use it, let's call for convenience's sake $t_{2}=2.5, d_{2}=1.5$ (instead of using t1 and d1, they are, after all just labels I can assign whichever meaning I want to!)
Thus, plugging into the formula, we have:
$$\frac{d_{1}}{1.5}=(\frac{t_{1}}{2.5})^{2}$$
Now, the next instant happens at 2.5 seconds afterwards, that is at time 2*2.5=5 seconds.

Let's plug this value into t1's place, and we get:
$$\frac{d_{1}}{1.5}=(2)^{2}=4$$
or multiplying with 1.5 both sides:
$$d_{1}=6.0$$
that is, the distance associated with time 5 seconds is 6 meters. Do you get that?

 

[bob1182006]

Well, but that is the beauty of the equation!

To show you how to use it, let's call for convenience's sake $t_{2}=2.5, d_{2}=1.5$ (instead of using t1 and d1, they are, after all just labels I can assign whichever meaning I want to!)
Thus, plugging into the formula, we have:
$$\frac{d_{1}}{1.5}=(\frac{t_{1}}{2.5})^{2}$$
Now, the next instant happens at 2.5 seconds afterwards, that is at time 2*2.5=5 seconds.

Let's plug this value into t1's place, and we get:
$$\frac{d_{1}}{1.5}=(2)^{2}=4$$
or multiplying with 1.5 both sides:
$$d_{1}=6.0$$
that is, the distance associated with time 5 seconds is 6 meters. Do you get that?

just wanted to point out that $t_{1}=2.5$

 

[Cole07]

But where do you get the (2) in d1/1.5=(2)^2

 

[bob1182006]

Cole07 he got it by taking 5/2.5

but since $t_{1}=2.5$ t2 is actually $t_{2}=2.5*2=5$ so you can just plug in those numbers into the equation and solver for $d_{1}$

 

[Cole07]

so d1=2.34375 is this the first mark?

 

[bob1182006]

could you show your work? i got a different answer

 

[Cole07]

i took d1/1.5=(2.5/5)^2 and (2.5/5)^2=0.25 and then i multiplyed by 1.5 and i have found that mistake i think i have finally fried my brain for the fourth mark do i take 2.5 and multiply by 4 and use the same equation?

 

[bob1182006]

to find the fourth mark you'd take 1 set of d and t that you already know, you can use the second mark/time, and yes just plug the numbers in but remember that the time for the 4th mark will be 2.5*4

 

[arildno]

just wanted to point out that $t_{1}=2.5$

Sorry about that, thought the values were corresponding. You are right of course.

 

[Cole07]

Thank You Very Much it worked