Why does a person on a bus move to the back when the bus accelerates?

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Homework Help Overview

The discussion revolves around the concept of inertia as it relates to a person on a bus during acceleration. The original poster describes a scenario involving a cartoon illustrating how a person appears to move to the back of the bus when it accelerates, raising questions about the forces at play and the nature of inertia.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between inertia and motion, questioning whether the bus's acceleration is sufficient to overcome the person's inertia. They discuss the implications of inertia in various scenarios, including car crashes and the effects of forces during acceleration and braking.

Discussion Status

Participants are actively engaging with the concepts of inertia and momentum, with some providing clarifications and others expressing confusion. There is an ongoing exploration of the definitions and implications of inertia, as well as how it relates to motion in different contexts.

Contextual Notes

Some participants note confusion regarding the distinction between mass and weight, and how these concepts apply to inertia. There is also mention of the need for further understanding of momentum, which has not yet been covered in detail by the original poster.

Peter G.
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I was reading about inertia and there was this cartoon on my textbook.

This guy puts on rollerblades and steps on to a bus. "Magically" as the car accelerates he ends up at the end of the vehicle and when he wants to step out the bus he simply stands up and, "magically" ends up at the entrance as the bus brakes.

So when the bus accelerates it moves forward, the guy ends up on the back of the bus: Is this because the force of the bus acceleration is not great enough to overcome his inertia so as the bus moves forward he, without "moving", ends up at the back of the bus? Or is the bus exerting a force on him making him move?

I don't know, I mean they explained about the car crashes too and I understood that our bodies are moving and if no resultant force great enough to overcome our reluctance of keep moving forward acts on us we will keep on moving in a straight line and how inertia is dependent on weight, not on mass but I can't grasp the inertia concept on this diagram.

Thanks in advance
 
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Hi Peter! :smile:
Peter G. said:
So when the bus accelerates it moves forward, the guy ends up on the back of the bus:

Is this because the force of the bus acceleration is not great enough to overcome his inertia so as the bus moves forward he, without "moving", ends up at the back of the bus?

Or is the bus exerting a force on him making him move?

Forget force.

There is no force!

If he starts to stand on the rollers when the bus is moving at 30 mph, then he always keeps moving at 30 mph, whatever the bus does…

i] if the bus keeps moving at 30 mph, then he stays where he is on the bus simply because they're moving at the same speed
ii] if the bus brakes, then he starts overtaking the bus (he's always going at 30 mph)
iii] if the bus accelerates, then the bus starts overtaking him (he's always going at 30 mph) :wink:
… how inertia is dependent on weight, not on mass …

This is very confusing :redface:

even if they mean it the other way round, the difference really only applies on the Moon etc

also, they're confusing two meanings of "inertia" … in your bus cartoon, it means the tendency to keep going, and this has nothing to do with either mass or weight … in a car crash, it means your momentum , which governs the forces on you when your speed is reduced.
 


Your explanation is really good. :smile:

Yeah, the mass and weight thing is confusing but they used it to explain that if you kick a rock in the moon it will still hurt :biggrin:

Regarding the car crash I think they also mean the tendency of keep moving because as soon as the car crashes it stops moving but no resultant force acts on you (provided you are not wearing your seat bealt) and you will keep moving through the windshield. I haven't learned about momentum, so I don't understand how it relates, maybe if you can explain o:).

Now, back to the bus: You said that when the bus moves at 30 mph, the man will move at 30 mph and, because of inertia, when the bus changes its motion, he does not, he continues moving the same way as he was before until another force acts upon him.

But he hops in the bus when it is at 0 mph, when the bus accelerates he continues moving at 0 mph? Or he accelerates together with the bus, because if he does, he will be moving at the same speed as the bus and will not end up at the back of the bus, right?

Sorry, I am a bit confused.

Thanks for your patience.
 
Hi Peter! :smile:

(just got up :zzz: …)
Peter G. said:
Yeah, the mass and weight thing is confusing but they used it to explain that if you kick a rock in the moon it will still hurt :biggrin:

Yes, but then it is the mass, not the weight! :wink:
Regarding the car crash I think they also mean the tendency of keep moving because as soon as the car crashes it stops moving but no resultant force acts on you (provided you are not wearing your seat bealt) and you will keep moving through the windshield. I haven't learned about momentum, so I don't understand how it relates, maybe if you can explain o:).

uh-uh, leave that until your teacher tells you about Newton's second law and conservation of momentum.
Now, back to the bus: You said that when the bus moves at 30 mph, the man will move at 30 mph and, because of inertia, when the bus changes its motion, he does not, he continues moving the same way as he was before until another force acts upon him.

But he hops in the bus when it is at 0 mph, when the bus accelerates he continues moving at 0 mph? Or he accelerates together with the bus, because if he does, he will be moving at the same speed as the bus and will not end up at the back of the bus, right?

If he's on the rollers when the bus starts, he'll hit the back of the bus.

If he's not (I don't know what the technical term is :redface:), then the friction between his shoes and the bus floor will be a force that will accelerate him, keeping him at the same speed (and acceleration) as the bus. :smile:
 


Ok, got it :wink:

Now, because I'm a very curious boy... :shy: I decided to look into momentum (there were only two pages anyway :biggrin:) And I was wondering if you could read two simple problems and say either: Correct or Wrong. In case I get it wrong, I promise I will sit, patiently, waiting for my lesson.

Force a car exerts on a wall during a crash: (Sorry I love cars)

So the car, of mass 200 kg, is traveling down the road at a velocity of 10 m/s. Suddenly, he hits a very rigid wall:

So, I will use the equation derived from Newton's Second Law of Motion: Ft = mv-mu

Ft = (200 x 10) - (200 x 0) (Quite a large Kinetic Energy dissipation :eek:)
Ft = 2000 kg m/s
(For the time of the collision, I will use the figure given in the book for a different problem involving a very rigid stone)
F = 2000 divided by 1/100th of a second
F = The car exerts on the wall is 200 000 N

The second problem involves the theory of conservation of momentum:

This time the car hits a person and I am wondering at what speed will this poor little guy move once struck by this misgoverned automobile:

The car, again, weighs 200 kg and is moving at 10 m/s, momentum of: 2000 kg m/s
The guy had a mass of 10 kg.

So: 2000 kg m/s = 10 kg x Velocity
Therefore: 2000 / 10 = 200 m/s

Thanks once again,
Peter
 
Hi Peter! :smile:

Your first example is ok
Peter G. said:
The second problem involves the theory of conservation of momentum:

This time the car hits a person and I am wondering at what speed will this poor little guy move once struck by this misgoverned automobile:

The car, again, weighs 200 kg and is moving at 10 m/s, momentum of: 2000 kg m/s
The guy had a mass of 10 kg.

So: 2000 kg m/s = 10 kg x Velocity
Therefore: 2000 / 10 = 200 m/s

No, momentum is conserved (in any collision), but the momentum of the car is not entirely transferred to the guy.

He and the car will normally have the same speed after the collision (in practice, the car would probably carry on braking, so the guy would fly off) …

so you can use the equation vcar,after = vguy,after, and combine it with the conservation of momentum equation. :wink:
 

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