Basic circuit/intensity and distance vs time graph question.

[Griffy11]

Hi everyone,

First question:

A radio-controlled car on a straight track accelerates uniformly from rest to 1.5m/s-1 in 5 seconds. The car continues at a constant velocityh for 10 sec, and decelerates uniformly to come to rest in 2 sec.

a) Draw a velocity-time graph for the data above. -Done.
b) Calculate the acceleration at 3 seconds after start:

a=v-u/t
a=0.9-0/3
a=0.3m/s

Is that right?

c) Calculate total distance travelled.
Here, I have no idea. I thought distance=speed/time at first, but that makes no sense at all. Then I thought displacement:

r=ut+1/2at^2

Would that be the right formula to use to find the total distance travelled, or am I doing something wrong?

Second thing, which seems basic... (sorry for the crappy diagram)

http://img62.imageshack.us/img62/383/quickdiagram.jpg

Compare the effect on the intensity of the light globe when another identical globe is added in series as opposed to adding it in parallel.

What I thought was that in series, both lights would be quite dim, but in parallel... heck I'm rambling, I have no real idea. =\

Thanks for the help in advance everyone, it's greatly appreciated. :)

 

[rl.bhat]

You have drawn the graph. It is trapezium. Find the area of that. That is the required result.
Illumination depends on I^2*R.
When the two bulbs are in series, the resistance is more. For a given voltage, the current in the circuit is less. So the illumination is less.
When the two bulbs are connected parallel and connected them across the same voltage, what will be the net resistance and current in the circuit?

 

[arithmetix]

The lights will both be dimmer than was the first light running by itself. This is because there is an'equivalent resistance' in the dry cell which tends to limit current.

As for the dynamics stuff, it looks pretty good to me.

 

[Griffy11]

You have drawn the graph. It is trapezium. Find the area of that. That is the required result.
Illumination depends on I^2*R.
When the two bulbs are in series, the resistance is more. For a given voltage, the current in the circuit is less. So the illumination is less.
When the two bulbs are connected parallel and connected them across the same voltage, what will be the net resistance and current in the circuit?

Great, found the area of the trapezium using 1/2(a+b)h

a=1/2(a+b)h
a=1/2(17+10)*1.5
a=20.25 units²

Two bulbs in series = greater resistance = dimmer light. (like you said)
Two bulbs in parallel, same voltage, the net resistance would increase, decreasing current again? So the light is dimmer in the parallel circuit as well as in series?

Hopefully that's right - thanks for the help!

 

[rl.bhat]

Area under the curve is the distance traveled, not the acceleration.
When two bulbs are in parallel net resistance decreases.So current increases for same voltage.

 

[Griffy11]

Area under the curve is the distance traveled, not the acceleration.
When two bulbs are in parallel net resistance decreases.So current increases for same voltage.

Yeah, I used a to define area, probably should have used d.

Also, with the area, I got 20.25 using the 1/2(a+b)h formula, but got 19.875 using the area of the two triangles then the area of the rectangle. Which way is correct?

To sum up the light:
Series - resistance increases, so current decreases, intensity decreases.
Parallel - resistance increases, so current decreases, so again, intensity decreases.
Therefore both bulbs get dimmer.

 

[rl.bhat]

20.25 m is the correct answer.
If R1 and R2 are connected in parallel, what is the equivalent resistance? Whether it is more then, equal to or less than series combination?

 

[Griffy11]

If R1 and R2 are connected in parallel, the equivalent resistance would be lower than the series combination... I think.

 

[rl.bhat]

Yes. Therefore in the parallel combination intensity increases.

 

[Griffy11]

Great - thanks a lot for all your help, I appreciate it a lot! :)