B Very basic acceleration problem (from a video on The Physics Classroom website)

[paulb203]

I'm confused after 2 minutes of this video on acceleration;

It starts with telling us that a car acclerates at 8m/s/s for 5 seconds.
Then it gives a data chart which includes the car's position at each time interval. The data is as follows.
O seconds; O metres
1 second; 4 metres
2 seconds; 16 metres
3 seconds; 36 metres
4 seconds; 64 metres
5 seconds; 100 metres

I can see the acceleration of 8m/s for the following intervals; 1-2 seconds; 2-3 seconds; 3-4 seconds; and 4-5 seconds.

But for the first second (0-1 seconds) it looks to me like an acceleration of only 4m/s (the car starts at 0m/s and acclerates to 4m/s).
Where am I going wrong?

 

[Ibix]

If it did a constant 4m/s for 1s, how far would it travel?

If it accelerated from rest to 4m/s in 1s, would it travel the same distance, or less, or more? What does that tell you about the plausibility of your "accelerated to 4m/s" idea?

 

[paulb203]

Thanks, Ibix

“If it did a constant 4m/s for 1s, how far would it travel?”

4m?

“If it accelerated from rest to 4m/s in 1s, would it travel the same distance, or less, or more?”

Less? It would be travelling at less than 4m/s for some of that second?

“What does that tell you about the plausibility of your "accelerated to 4m/s" idea?”

That it has zero plausibility ). Thanks.

How does going from 0m/s to 4m/s amount to an acceleration of 8m/s though?

Or is that not what happened? Is 4m/s the average velocity for that 1st second?

Is it going at 8m/s by the END of that first second?

 

[kuruman]

You deduced that the car has velocity 4 m/s by dividing the 4 m it traveled by the interval of 1 second. The formula "speed equals distance divided by time" does not work when the object is accelerating. In fact, when the acceleration is constant, d Distance traveled in a time interval divided by the time interval is the average velocity over that interval. When the acceleration is constant, it is equal to the instantaneous velocity at the midpoint of the time interval. In this case, the instantaneous velocity of the car is 4 m/s at t = 0.5 s and 8 m/s at t = 1 s.

On edit: @Ibix's point in post #5 is well taken. The misplaced qualifier "when the acceleration is constant" has been moved to its rightful place.

 

[Ibix]

In fact, when the acceleration is constant, distance traveled in a time interval is the average velocity over that interval.

I think you mis-spoke slightly here - this is true whether you have constant acceleration or not. But the average velocity is only equal to the average of the initial and final velocities if the acceleration is constant.

 

[kuruman]

Yes, it can be misleading. I edited the post to fix the problem.

 

[paulb203]

Thanks, kuruman.

“When the acceleration is constant, it is equal to the instantaneous velocity at the midpoint of the time interval. In this case, the instantaneous velocity of the car is 4 m/s at t = 0.5 s and 8 m/s at t = 1 s.”

This is helpful. I think I inadvertently worked out that 4m/s was the instantaneous velocity at the midpoint (0.5s).

I was thinking that if it starts at 0m/s, and ends up at 8m/s, over 1s, and the acceleration is constant, the increases in velocity might go something like;

Os; 0m/s

1/8s; 1m/s

2/8s; 2m/s

3/8s; 3m/s

4/8s; 4m/s (the midpoint you referred to)

5/8s; 5m/s

6/8s; 6m/s

7/8s; 7m/s

8/8s; 8m/s

Is this correct?

 

[PeroK]

Is this correct?

Yes.

 

[paulb203]

Thanks, PeroK

 

[paulb203]

I noticed that the velocities given in the video only referred to magnitude. And I've seen this before in textbook examples. Yet velocity, like acceleration is, I'm told, a vector quantity. If the direction of the velocity (or the acceleration) isn't relevant, and the video/textbook is only concerned with magnitude, do they always leave out the direction? And is it technically ok to do that? Can we correctly refer to velocity and acceleration in terms of magnitude alone?

 

[Lnewqban]

@matthewphilip , please see:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/motgraph.html#c1

As you can see, a velocity versus time graph for constant acceleration will have the shape of a triangle.
The area below the line is always the covered distance in that time.

 

[PeroK]

I noticed that the velocities given in the video only referred to magnitude.

I didn't want to watch that whole video, but the introduction mentioned velocity having magnitude and direction.

And I've seen this before in textbook examples. Yet velocity, like acceleration is, I'm told, a vector quantity.

It definitely is.

If the direction of the velocity (or the acceleration) isn't relevant, and the video/textbook is only concerned with magnitude, do they always leave out the direction?

They don't (always leave out direction).

And is it technically ok to do that? Can we correctly refer to velocity and acceleration in terms of magnitude alone?

No. In one dimension the direction reduces to a simple $\pm$. With positive velocity values being in one direction and negative values being in the other. In two or three dimensions, the full vector nature is seen, with a velocity have two or three components.

 

[paulb203]

Thanks again, PeroK

"In one dimension the direction reduces to a simple ±. With positive velocity values being in one direction and negative values being in the other."

Always positive to the right, negative to the left, like on a number line?

 

[PeroK]

Always positive to the right, negative to the left, like on a number line?

You are free to choose. You can have positive to the left. More importantly, positive and negative could be up and down. This is where it is important to understand your convention. If up is positive, then the acceleration due to gravity is negative: $a = -g = -9.8 m/s^2$. And, of course, if down is positive, then $a = g = 9.8 m/s^2$.

You must always get the same physical answer, regardless of the up/down or left/right convention you've chosen.

 

[jbriggs444]

Thanks again, PeroK

"In one dimension the direction reduces to a simple ±. With positive velocity values being in one direction and negative values being in the other."

Always positive to the right, negative to the left, like on a number line?

That is the usual convention. But it is only a convention. We usually call it a "sign convention".

You can solve a problem using a convention of positive for right and negative for left. Or you can use the opposite convention. You will get the same answer either way -- bearing in mind that a positive result using the one sign convention is the "same answer" as a negative result using the other.

You can also use positive up/negative down. Positive north, negative south. Or their reverses. Any convention is valid, as long as you identify it and consistently adhere to it. [And as long as you report your final results using the convention that the instructor expects].

Edit: Scooped by @PeroK. Drat him!

 

[kuruman]

I noticed that the velocities given in the video only referred to magnitude. And I've seen this before in textbook examples. Yet velocity, like acceleration is, I'm told, a vector quantity. If the direction of the velocity (or the acceleration) isn't relevant, and the video/textbook is only concerned with magnitude, do they always leave out the direction? And is it technically ok to do that? Can we correctly refer to velocity and acceleration in terms of magnitude alone?

The directions of the velocity and the acceleration vectors are very relevant. Consider a ball thrown straight up with velocity vector v = +20 m/s. The acceleration of gravity is (roughly) -10 m/s², negative because it points opposite to the velocity which we chose to be positive.

Now for every second that goes by, we add -10 m/s to the velocity that is already there. Look at the statements below. The first in parentheses is the velocity that is already there, the second term is the constant velocity increment that is added.
After 1 s the velocity is (20) + (-10) = 10 m/s i.e. the ball is moving up and the speed is decreasing.
After the next second the velocity is (10) + (-10) = 0 m/s i.e. the ball is instantaneously at rest and the speed is zero.
After the next second the velocity is (0) + (-10) = -10 m/s i.e. the ball is moving down and the speed is increasing.
After the next second the velocity is (-10) + (-10) = -20 m/s i.e. the ball is moving down and the speed is increasing. At this time, 4 seconds from launch, the ball has returned to the launching point with the same speed that it had but in the opposite direction. What goes up must come down.

So the relative between the velocity and the acceleration is very important:

1. If the acceleration is in the same direction as the velocity, the speed (magnitude of the velocity) is increasing.
2. If the acceleration is in the opposite direction as the velocity, the speed is decreasing.
3. If the acceleration is in the perpendicular to the velocity, the speed stays the same but the velocity changes direction (the object turns).

You can combine the ideas above. If the acceleration is at an arbitrary angle relative to the velocity, then the speed will increase or decrease while the object will turn. If you think about it, driving a car involves adjustment of the angle between acceleration and velocity. The accelerator pedal provides an acceleration component in the same direction as the velocity; the brake pedal provides an acceleration component opposite to the velocity; the steering wheel provides an acceleration component perpendicular to the velocity either to the left or to the right. So you see, the relative direction of acceleration and velocity is quite important for controlling the motion of anything that moves.

 

[jbriggs444]

The acceleration of gravity is (roughly) g = -10 m/s2

The usual convention is that "g" is the always positive magnitude and that the vector acceleration value depends on one's chosen sign convention. So the acceleration of a freely falling object under an "upward positive" convention is roughly $-g = -10 m/s^2$

 

[kuruman]

The usual convention is that "g" is the always positive magnitude and that the vector acceleration value depends on one's chosen sign convention. So the acceleration of a freely falling object under an "upward positive" convention is roughly $-g = -10 m/s^2$

Far be it from me to add to the confusion about whether $g$ is positive or negative. I edited the post to remove the offending symbol. Thanks.

 

[Mark44]

I can see the acceleration of 8m/s

But for the first second (0-1 seconds) it looks to me like an acceleration of only 4m/s

I haven't noticed that anyone else commented on the units here. Acceleration is always in units of $\frac{\text{distance}}{\text{time}^2}$. The m/s unit would be a velocity.

O seconds; O metres
1 second; 4 metres

Unless your keyboard is very different from mine, it looks like you have typed the letter 'O' instead of the numeric digit '0'.

 

[kuruman]

I haven't noticed that anyone else commented on the units here. Acceleration is always in units of distancetime2. The units you showed are those or velocity.

I think it's a typo. OP seems to be aware of the correct units. Note the first line in post #1.

It starts with telling us that a car acclerates at 8m/s/s for 5 seconds.
Then it gives a data chart which includes the car's position at each time interval. The data is as follows.
O seconds; O metres
1 second; 4 metres
2 seconds; 16 metres
3 seconds; 36 metres
4 seconds; 64 metres
5 seconds; 100 metres

I can see the acceleration of 8m/s for the following intervals; 1-2 seconds; 2-3 seconds; 3-4 seconds; and 4-5 seconds.

 

[Mark44]

I think it's a typo.

A very consistent one, though, as $\frac m {s^2}$ isn't used again after the first time.

 

[kuruman]

A very consistent one, though, as $\frac m {s^2}$ isn't used again after the first time.

Maybe consistent, maybe not. Post #7 shows that OP understands the difference between velocity and acceleration and the idea that the acceleration in this case is the "speed of the speed".