Phyiscs kinematics question

[jammin06]

---

Members do not see this ad.

sup everyone,
generally i don't have any problem w/ kinematics on the physics section of the exam, but there's one problem in the PR science workbook that has got me. anyone got any advice on this one?

Question: One second after being thrown straight down, and object is traveling at a speed of 20 m/s. How fast will it be traveling two seconds later?

a) 30 m/s
b) 40 m/s
c) 50 m/s
d) 60 m/s

 

[stoleyerscrubz]

acceleration being constant: it should be moving at 60 m/s at 3 seconds.

i'm cramming for physics portion of the MCAT yesterday,today, and tommorow so my brain is a bit fried now.

 

[jammin06]

yes, that was what i put, but it's apparently incorrect. The correct answer is B, 40 m/s. this is the explination the book gives:

Using equation (delta)V = a(delta)t. we find the object gains: (delta) v = (10m/s2)(2s) = 20 m/s.....Its new speed is therefore Vf = vo + (delta) v = 20 + 20 = 40 m/s

I think this explination is whack. any takers?

 

[chopper]

NO, NO, NO.

Don't get fooled by the 'one second after it is being thown down' b.s. this is extraneous information and not important to the problem. How it started out getting to where it is in unimportant. All you need to know is that when you start, vel is 20m/s DOWN.

Solve this way:
Initial velocity is 20m/s down.
accel is 10m/s2 down.
what is velocity after 2 seconds?

v=v0 + at. 20 + 10(2) = 40 m/s.

 

[ih8biochem]

only force acting on the object is gravity (i.e. 10 m/s^2)
If the initial velocity is 20 m/s, than after one second, it's going 30 m/s.
After 2 seconds, it's going 40 m/s, and so on. So simple...

 

[Csv321]

NO, NO, NO.

Don't get fooled by the 'one second after it is being thown down' b.s. this is extraneous information and not important to the problem. How it started out getting to where it is in unimportant. All you need to know is that when you start, vel is 20m/s DOWN.

Solve this way:
Initial velocity is 20m/s down.
accel is 10m/s2 down.
what is velocity after 2 seconds?

v=v0 + at. 20 + 10(2) = 40 m/s.

Wow....I hate that extraneous info crap....I need to review my physics....You're a genius! 😀

only force acting on the object is gravity (i.e. 10 m/s^2)
If the initial velocity is 20 m/s, than after one second, it's going 30 m/s.
After 2 seconds, it's going 40 m/s, and so on. So simple....

I like that explanation too....And boo to biochem

 

[elias514]

The responses to this question assume that the acceleration is not constant--i.e., that the acceleration instantaneously changes at the one second mark from 20 m/s^2 to the acceleration due to gravity, approximately 10 m/s^2. I'm not sure that this assumption is valid. The only factor that would reduce the acceleration of the object in this case is friction, but it's impossible to infer anything about the role of friction here (is it negligible? is the object falling in a vacuum? what's the environment?).

Granted, one could argue that the object, in spite of the fact that it was thrown, actually had an initial velocity prior to the throwing. This would cast the notion that the initial acceleration is 20 m/s^2 into question; even so, the issue of acceleration NOT being constant remains.

If I were you, I'd forget this problem because the question stem is flawed.

 

[ih8biochem]

The responses to this question assume that the acceleration is not constant--i.e., that the acceleration instantaneously changes at the one second mark from 20 m/s^2 to the acceleration due to gravity, approximately 10 m/s^2. I'm not sure that this assumption is valid. The only factor that would reduce the acceleration of the object in this case is friction, but it's impossible to infer anything about the role of friction here (is it negligible? is the object falling in a vacuum? what's the environment?).

Granted, one could argue that the object, in spite of the fact that it was thrown, actually had an initial velocity prior to the throwing. This would cast the notion that the initial acceleration is 20 m/s^2 into question; even so, the issue of acceleration NOT being constant remains.

If I were you, I'd forget this problem because the question stem is flawed.

Keep in mind, F = ma. In order for acceleration to be greater than gravity, addtional force must be applied throughout the fall. Not true. No addtional force is applied besides gravity, and that's constant. Therefore, a is constant. All that's important is that v0 is 20 m/s.

You know, when I took Kaplan, there was always somebody there to telling people the question stem was flawed when they can't answer the problem. So dumb. Like you really understand physics better than the test writers? My rant.

 

[xlithiumx]

This problem is not as hard as you are making it, elias514.

Let's look at this question second by second.

Assume that timing starts when the object leaves the thrower's hand:

at t=0, the object has some initial velocity.
at t=1, the object is at 20 m/s. Since acceleration due to gravity is roughly 10 m/s^2, we can infer that the object's initial velocity was 10 m/s.
at t=2, the object is at 30 m/s
at t=3, the object is at 40 m/s.

The acceleration never changes. The object has an initial velocity of 10 m/s and increases velocity by 10 m/s each second.

In all things simplify simplify simplify.

 

[chopper]

For everyone: trying to read too much into MCAT questions is going to KILL you on the test. Stop thinking like they are trying to trick you about air resistance or non-constant accelration or whatever. They don't NEED to make it hard because they know enough people will go ahead and take care of that for them by not just doing the easy thing.

ELIAS: The acceleration NEVER changed. It was always 10 m/s2. Period. The VELOCITY was 20 m/s after 1 sec. The MCAT will NEVER give you a 'air resistance' problem UNLESS it is the theme of the passage (i.e. all they talk about is air resistance, and giving a bunch of equations for it). In this case, you would use the information given in the passage. And btw, the question stem DID say that air resistance could be neglected, so that was taken care of.

Bottom line: If you want to do well on the MCAT, stop trying to make each problem as god-awful difficult as you can, and using fuzzy logic to try to explain why these questions are 'wrong'. 999/1000 - you are wrong - BIOCHEM has a great point here.

Here is the formula for getting Kinematics problems right EVERY time for the MCAT:
1. Analyze info given. They will ALWAYS give you 3 of the 5 variables you need to answer ( those being a,d,t,v0, and vf). Remember some ways they give you 'hidden' initial variables: 'Starting from rest' means v0 = 0 m/s. If they talk about gravity, a can be assumed to be 10. "Until the ball stops going up" might mean vf = 0. Things like that.
2. Set up your reference frame. This means if you have initial velocity 'up', it and acceleration have different signs. If initial velocity is 'down', it has same sign as acceleration.
3. Sometimes you have to do in 2 parts. Like for the 'how long does a ball travel' questions. First you have to get time up/time down (using v0, vf, and accel) in Y direction. Then you go and plug in for distance in X direction.
4. Plug and chug (usually VERY easy numbers) into your equation to get what they want. Princeton has '5 equations' for this. I'm sure Kaplan and the others have something similar.

Good luck everyone. This stuff shouldn't be hard. If you find yourself saying 'well, I got this. maybe if the air resistance is taken into account . . . . . . ' - THEN STOP. You are doing something wrong. Do it over and see if it works.

chopper