"Help me, I'm thick as a brick." [EK 1001 PhysicsQ Thread]

apoptos · Jul 16, 2007

50). A particle starts from rest and travels in a straight line for 4s.
If the particle is accelerating at a constant rate, which of the following could be the distances travelled by the particle during each consecutive second?

A. 10m, 20m, 30m, 40m
B. 5m, 15m, 25m, 35m
C. 5m, 25m, 125m, 625m
D. 2m, 4m, 8m, 16m

How should I approach this problem? I don't think I fully grasp the answer given.

Kaydubz · Jul 16, 2007

apoptos said:
50). A particle starts from rest and travels in a straight line for 4s.
If the particle is accelerating at a constant rate, which of the following could be the distances travelled by the particle during each consecutive second?

A. 10m, 20m, 30m, 40m
B. 5m, 15m, 25m, 35m
C. 5m, 25m, 125m, 625m
D. 2m, 4m, 8m, 16m

How should I approach this problem? I don't think I fully grasp the answer given.

It's not often I can answer the same question twice! Digs through old posts...

Use the kinematic equations.

x = x0 + x0t + (1/2)at^2
simplifies to

x = (1/2)at^2

Now substitute time (t) = 1, 2, 3, and 4 into that equation.

So at 1 second, x = 0.5a
at 2 seconds, x = 2a
at 3 seconds, x = 4.5a
at 4 seconds, x = 8a

So then you can subtract (t-1) from (t) to find the distance traveled in:
the first second (t=0 to t=1): 0.5a
the second second (t=1 to t=2): 1.5a
the third second (t=2 to t=3) : 2.5a
the fourth second (t=3 to t=4) : 3.5a

Answer B is the only choice that fits the ratio 0.5 : 1.5 : 2.5 : 3.5

That didn't take as long as it looks btw

Tussis · Jul 16, 2007

Note the keywords within the question (which is especially important for the MCAT - those minute detail words that make a big difference). The question tells us 1: the object is starting from rest; and 2: the object has a constant acceleration.
So, for all answer choices, the given distance at time t=0 is Zero meters. So, one calculates the acceleration between 0m and the first distance for each answer choice, with the time difference being 1 second.So, using the quick calculations with the kinematics equation x(f)=1/2(a)t^2 [assuming x(o)=0 and v(o)=0], we come up with the acceleration for A thru D being:
A = 20 m/s^2
B = 10 m/s^2
C = 10 m/s^2
D = 4 m/2^2 [Notice the a is just double the first distance.]

Finally,find the total distance traveled by taking your Acceleration for each, and plugging into the equation above, so in 4 seconds, A has gone 160m, B = 100m, C = 100m, and D = 32m.

Now, add the given distances up to check which one totals the correct distance. You'll find that answer B is the only one that correctly matches.
[You should've eliminated C already due to it being preposterously out of whack.]

EDIT: Or the above method.

apoptos · Jul 16, 2007

Kaydubz said:
It's not often I can answer the same question twice! Digs through old posts...

Thanks to you both...
And don't worry, I'll have new ones for you soon!

apoptos · Jul 17, 2007

[EDIT] nvm, figured it out. lol

Bartelby · Jul 18, 2007

apoptos said:
50). A particle starts from rest and travels in a straight line for 4s.
If the particle is accelerating at a constant rate, which of the following could be the distances travelled by the particle during each consecutive second?

A. 10m, 20m, 30m, 40m
B. 5m, 15m, 25m, 35m
C. 5m, 25m, 125m, 625m
D. 2m, 4m, 8m, 16m

How should I approach this problem? I don't think I fully grasp the answer given.

This is a good example of where thinking it through can be easier than doing any calculations. You know acceleration is constant, so it will have to increase the distance it is covering by a constant amount each time. That gets rid of C, D, and B (for B it increased from 0 to 5 in second one but 5 to 15 in second two, 5 vs. 10). That leaves you with A. Figuring out some simple approaches like these can save you a lot of time on test day.

KeyzerSoze · Jul 18, 2007

Bartelby said:
This is a good example of where thinking it through can be easier than doing any calculations. You know acceleration is constant, so it will have to increase the distance it is covering by a constant amount each time. That gets rid of C, D, and B (for B it increased from 0 to 5 in second one but 5 to 15 in second two, 5 vs. 10). That leaves you with A. Figuring out some simple approaches like these can save you a lot of time on test day.

I don't think that's correct. In choice A, during the first second, the average velocity is 10 m/sec. The particle started from rest, so at the end of the first second, it must be going 20 m/s, to give you an average of 10 m/s. Therefore the acceleration is 20 m/s^2. In the 2nd second, the average velocity is 20 m/s, which would give no acceleration at all. Therefore the particle is not accelerating at a constant rate.

In choice B, during the first second, the average velocity is 5 m/sec. The particle started from rest, so at the end of the first second, it must be going 10 m/s, to give you an average of 5 m/s. Therefore the acceleration is 10 m/s^2. In the 2nd second, the average velocity is 15 m/s. We know it started the 2nd second with a velocity of 10 m/s, so at the end of the 2nd second, it must be going 20 m/s, to give you an average of 15 m/s. The acceleration remains constant at 10 m/s^2, and choice B is the correct choice.

This is an example of where a choice is written to look correct, to get you to think it through incorrectly. 🙂

Bartelby · Jul 19, 2007

KeyzerSoze said:
I don't think that's correct. In choice A, during the first second, the average velocity is 10 m/sec. The particle started from rest, so at the end of the first second, it must be going 20 m/s, to give you an average of 10 m/s. Therefore the acceleration is 20 m/s^2. In the 2nd second, the average velocity is 20 m/s, which would give no acceleration at all. Therefore the particle is not accelerating at a constant rate.

In choice B, during the first second, the average velocity is 5 m/sec. The particle started from rest, so at the end of the first second, it must be going 10 m/s, to give you an average of 5 m/s. Therefore the acceleration is 10 m/s^2. In the 2nd second, the average velocity is 15 m/s. We know it started the 2nd second with a velocity of 10 m/s, so at the end of the 2nd second, it must be going 20 m/s, to give you an average of 15 m/s. The acceleration remains constant at 10 m/s^2, and choice B is the correct choice.

This is an example of where a choice is written to look correct, to get you to think it through incorrectly. 🙂

Hehe, guess I should have given a bit more thought to this one. At any rate, my method DID save time :laugh:

.

apoptos · Jul 26, 2007

392). If the two masses stick together after colliding, what is the final horizontal velocity of the masses?

A. 0 m/s
B. 5 m/s to the left
C. 7 m/s to the left
D. 9 m/s to the left

It says the answer is C. How is the answer C? I have D.
[EDIT] Nevermind. I was adding instead of subtracting. Got it now. Hate physics.

apoptos · Jul 26, 2007

285). Boxes A, B and C each have a mass of 10kg. If the boxes are accelerated upward at 2 m/s2 what is the tension T2?

A. 110 N
B. 220 N
C. 240 N
D. 360 N

Tan008 · Jul 26, 2007

apoptos said:
392). If the two masses stick together after colliding, what is the final horizontal velocity of the masses?

A. 0 m/s
B. 5 m/s to the left
C. 7 m/s to the left
D. 9 m/s to the left

It says the answer is C. How is the answer C? I have D.
[EDIT] Nevermind. I was adding instead of subtracting. Got it now. Hate physics.

the answer is C, but you have your diagram wrong...the larger sphere is 20kg, not 10 kg

eltor07 · Jul 26, 2007

apoptos said:
285). Boxes A, B and C each have a mass of 10kg. If the boxes are accelerated upward at 2 m/s2 what is the tension T2?

A. 110 N
B. 220 N
C. 240 N
D. 360 N

I think the answer is C... Both boxes under T2 are added together as a single weight and then you just do T-20(10)=20(2) and solve for T

apoptos · Aug 8, 2007

A 12-g object, dropped into a container of ethyl alcohol (specific gravity = .8) has an apparent mass of 8g.
What is the approximate density of the object?

a. 1.0 g/cm3
b. 1.6 g/cm3
c. 2.4 g/cm3
d. 5.0 g/cm3

They give solution to this one but I don't really follow the first half...
Thanks for all your help so far.

CremasterFlash · Aug 8, 2007

here's my guess:

since it displaces 4g (weight - apparent weight once submerged) and we know that the liquid has a mass of .8 g/cm^3 then we know how much liquid it displaces: 4/.8 which is 5 cm^3

since it displaces 5cm^3 and has a weight of 12gm then its density should be 12/5 g/cm^3 or 2.4

of course i've had a few beers so this could all be wrong.

Tan008 · Aug 8, 2007

I have a kinda complex way of doing the question, but that's how I was taught to do it in PHYS (this is the way Archemedes figured out whether the crown was made from pure gold or not 😀)

Apparent mass is what you would get if you attached a scale to your submerged object. This attached scale and the reading can be represented by tension (this tension force will be pointing in the same direction as the buoyant force). So the tension force would be the apparent weight of the object --> T= 8 x 10^-3 x 10.
Buoyant force is another force poiting up, and then you have mg pointing down. Just set the upward force components (tension and Fb) equal to the downward (mg) and solve for the volume of the object. Then you can use the volume to get density. This is a much longer way of doing this problem though.

you end up getting 2.4 if you solve through for density 😀

apoptos · Aug 9, 2007

You guys are the bomb.

A conducting wire is coneceted to both plates of a fully charged capacitor with a 3 nF capacitance and 3 mV voltage, the stored charge crosses teh wire in 4 microseconds. What is current across the wire?

A. 3.60 x 10^-17 A
B. 2.25 x 10^-6 A
C. 2.24 x 10^-1 A
D. 7.50 x 10^-1 A

I get that Q=VC = (3mV)(3nF) = (3x10^-3V)(3x10^-9F) = 9x10^-12C
But how do you get the current across the wire from here?

axp107 · Aug 9, 2007

apoptos said:
You guys are the bomb.

A conducting wire is coneceted to both plates of a fully charged capacitor with a 3 nF capacitance and 3 mV voltage, the stored charge crosses teh wire in 4 microseconds. What is current across the wire?

A. 3.60 x 10^-17 A
B. 2.25 x 10^-6 A
C. 2.24 x 10^-1 A
D. 7.50 x 10^-1 A

I get that Q=VC = (3mV)(3nF) = (3x10^-3V)(3x10^-9F) = 9x10^-12C
But how do you get the current across the wire from here?

Current = charge/time

You have to know that formula.. memorize it :luck:

"Help me, I'm thick as a brick." [EK 1001 PhysicsQ Thread]

apoptos

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Tussis

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