Understanding d=1/2gt^2 for Beginner Physics Students

[holly]

Can't Understand t**2

I hope I won't get blasted for asking for help on something everyone else in my introductory physics class can grasp: I cannot understand, in my heart, the d=1/2gt**2 formula when determining the distance a ball in free-fall travels in x seconds with the 1/2 g being 4.9. I recognize when to use it. I can parrot it. But I just can't grasp why the time is being squared, tho' I know that gives the right answer. It seems to me that a second is a second. I'm not able to get more seconds out of a second! The idea of meters per second I can grasp. But the idea of squaring the SECONDS I just can't get. I'm just a lowly sonography student taking physics. Any help on grasping why we're squaring time would be appreciated. I have tried and tried to figure it out.

 

[chroot]

The gravitational acceleration is 9.8 meters per square second.

The object falling gains 9.8 meters/sec of velocity in each second it falls.

Its velocity increases 9.8 meters per second... per second. The acceleration is thus 9.8 m/s/s, or 9.8 m/s^2.

Do you know calculus? There's also a clean explanation there: you differentiate position with respect to time once to get velocity. You differentiate it again with respect to time to get acceleration. Acceleration has to deal with second-order time quantities.

- Warren

 

[krab]

Originally posted by holly
I hope I won't get blasted for asking for help on something everyone else in my introductory physics class can grasp: ..

2 points:

1. You won't get blasted; this isn't that kind of forum.

2. You are the one asking the intelligent question. Don't believe that everyone else in your class can grasp it. Many students think that understanding means the same thing as being able to parrot a formula: they won't go far. You OTOH have a better definition of "understand", and that's the most important thing you need to know in order to learn physics.

Now to the question. The g gives the rate of increase of speed. (If it helps, imagine that the velocity unit is a vel, and 1 vel = 1 m/s. Then g = 9.8 vel/s.)

Now the trickier part. How to find distance traveled when velocity is steadily increasing. If a mass starts falling from rest, then after a time t, its speed is $v_f=gt$. Now imagine that it had gone the whole distance at that speed. Then the distance d is $d = v_f t = gt^2$. That's clearly an overestimate because since it starts with speed zero, its average speed is $v_a = 1/2gt$. So it makes sense that the distance traveled in time t is $d = v_a t = 1/2gt^2$. This isn't really a proof, but hopefully, I've made it at least believable.

 

[Razzy]

Great question! Please allow me the opportunity to explain quite simply. First, it is important to understand that when multiplying things (i.e. numbers), they are being distributed “throughout“. Now imagine the concept of “times squared“. This is similar to a fly you might see flying inside the bus in the same direction as the bus. Now let‘s assume the bus is traveling at a speed of 20km per hour. If the fly is moving in the same direction as the bus from the back of the bus to the front
of the bus, the fly is accelerating relative to the constant velocity of the bus. If the fly is also traveling towards the front of the bus at 50km per hour, it has essentially increased its speed 20km per hour throughout its journey from the end of the bus to the front. Therefore, squaring its speed relative to the road surface outside the bus. Sorry in advance if you can‘t understand. I am typing this on my mobile.

 

[JaredJames]

This thread is seven years old and the OP hasn't been back since that time. You need to check this before posting.

I don't see what you posted has to do with the OP. In fact, your post appears to be gibberish.

 

[KingNothing]

Maybe we should have a "Recognition" called "Thread Necromancer".

 

[JaredJames]

Maybe we should have a "Recognition" called "Thread Necromancer".

I fear people would deliberately go for it.

 

[Borek]

Locked, just in case someone want to add something more.