Relative acceleration and others

In summary, the customer slides an empty beer mug off the counter and it has an acceleration of -12.35 m/s2 relative to the counter.
  • #1
ppxrare
10
0
A bolt drops from the ceiling of a train car that is accelerating northward at a rate of 2.55 m/s2.
(a) What is the acceleration of the bolt relative to the train car? (they also want the angle to the south from the vertical)
(b) What is the acceleration of the bolt relative to the Earth?

So the book does not explain relative acceleration properly so I just did -9.8-2.55 = -12.35 for a) which I got wrong and for b) I said its -9.8 which was also wrong
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Here is a question which I got right as it was a yes or no and I had more than 1 submission in webassign,
(a) Can a particle moving with instantaneous speed 6.00 m/s on a path with radius of curvature 2.00 m have an acceleration of magnitude 29.00 m/s2?

(b) Can it have an acceleration of magnitude 17.00 m/s2?

At first I answered both no because I got the acceleration a=v^2/r 18m/s^2 and isn't the centripetal acceleration constant? the correct answers are yes and no , I don't understand though
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In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.38 m. The mug slides off the counter and strikes the floor 0.20 m from the base of the counter.
(a) With what velocity did the mug leave the counter?
0.376867


(b) What was the direction of the mug's velocity just before it hit the floor?° (below the horizontal)
274.140

So in webassign I got a) correct but b) wrong although I'm pretty sure its right , I do understand the question the Vxi is the answer in a) and the Vyi is -5.200769 and I got theta =-85.85537 they want below the horizontal so I just added 360 to it, Do they want a different angle? please help
 
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  • #2
welcome to pf!

hi ppxrare! welcome to pf! :smile:

(btw, it's not a good idea to put too many questions in the same thread :wink:)

question 1: acceleration is a vector, so (like velocity) it obeys the laws of vector addition (though -9.8 looks right to me :confused:)

question 2: ah, but centripetal acceleration isn't the same as total acceleration, is it? :smile:
 
  • #3
ppxrare said:
A bolt drops from the ceiling of a train car that is accelerating northward at a rate of 2.55 m/s2.
(a) What is the acceleration of the bolt relative to the train car? (they also want the angle to the south from the vertical)
(b) What is the acceleration of the bolt relative to the Earth?

So the book does not explain relative acceleration properly so I just did -9.8-2.55 = -12.35 for a) which I got wrong and for b) I said its -9.8 which was also wrong
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That's wrong because acceleration is a vector not a number! Since the bolt, once it has started dropping, maintains the same speed as the car but has no northward acceleration, the acceleration vector, relative to the train car, is a vector with downward acceleration that of gravity and southward acceleration the same magnitude as that of the train car. The acceleration relative to the Earth is a vector downward with magnitude 9.8.

Here is a question which I got right as it was a yes or no and I had morecao than 1 submission in webassign,
(a) Can a particle moving with instantaneous speed 6.00 m/s on a path with radius of curvature 2.00 m have an acceleration of magnitude 29.00 m/s2?

(b) Can it have an acceleration of magnitude 17.00 m/s2?

At first I answered both no because I got the acceleration a=v^2/r 18m/s^2 and isn't the centripetal acceleration constant? the correct answers are yes and no , I don't understand though
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Yes, it is correct that the acceleration perpendicular to the path (your "centripetal acceleration") is 18 m/s^2. Depending upon acceleration in the direction of the path, the magnitude of the acceleration can be any number larger than 18 m/s^2 but cannot be lower.

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.38 m. The mug slides off the counter and strikes the floor 0.20 m from the base of the counter.
(a) With what velocity did the mug leave the counter?
0.376867


(b) What was the direction of the mug's velocity just before it hit the floor?° (below the horizontal)
274.140

So in webassign I got a) correct but b) wrong although I'm pretty sure its right , I do understand the question the Vxi is the answer in a) and the Vyi is -5.200769 and I got theta =-85.85537 they want below the horizontal so I just added 360 to it, Do they want a different angle? please help
If you calculated -85.85537 as the angle, then the angle "below the horizontal" is +85.85537. Draw a picture to see that.
 
  • #4
Halls of Ivy, thanks a lot I got question 3 correct and I understood 2, as for 1) what will I choose for the direction for a) ?
 
Last edited:
  • #5
I still got a) for question 1 wrong, the acceleration and direction :( , can you please give me the answer as I know acceleration is a vector... but I did 9.8 - 2.55 and I still got the wrong answer, and what was the angle to the south from the vertical?
 
  • #6
it isn't 9.8 - 2.55, it's 9.8 down and 2.55 north
 
  • #7
hmm but that's not an option in webassign , they want an answer in m/s2 at a degree to the south from the vertical

I need the degree
 
  • #8
yes, you need the magnitude of this vector, and the angle :smile:
 
  • #9
so the magniture should be 7.25 downwards as I said 9.8-2.25 , but it was wrong

Can you please give me the answer because I'm getting confused , I said one thing and you replied that it is wrong then you give me a tip telling me to do the same thing I did at the start.

Thanks a lot for the help though
 
  • #10
ppxrare said:
so the magniture should be 7.25 downwards as I said

no!

what is the magnitude of a vector that's 9.8 in one direction, and 2.55 in a perpendicular direction?
 
  • #11
but its not perpendicular? one acceleration is upwards and the other is downwards ? they should cancel each other out?
 
  • #12
To clarify:

An object dropped in a train traveling at a constant speed and direction will travel in the same x-direction as the train, at the same speed as the train.

It will accelerate toward the floor of the train just as if it were dropped from the same height on solid ground.

So, since the train is accelerating, that means that the speed of the train is increasing even after the bolt is dropped. The bolt does not accelerate in the x-direction (the direction of the train's movement). However, since the train is accelerating (in the x-direction), the bolt will be decelerating (accelerating negatively) relative to the train.
 
  • #13
ppxrare said:
but its not perpendicular? one acceleration is upwards and the other is downwards ? they should cancel each other out?

ah, no …

one is northward (horizontal), and the other is downwards :wink:
 

Related to Relative acceleration and others

1. What is relative acceleration?

Relative acceleration is the difference in acceleration between two objects or reference frames. It is the rate at which the velocity of one object changes relative to the velocity of another object.

2. How is relative acceleration calculated?

Relative acceleration can be calculated by taking the difference in acceleration between two objects or reference frames, and dividing it by the time interval in which the change in acceleration occurs.

3. What is the difference between relative acceleration and absolute acceleration?

Relative acceleration takes into account the movement of one object in relation to another, while absolute acceleration only considers the acceleration of a single object without any external reference frame.

4. How does relative acceleration affect motion?

Relative acceleration can affect motion in many ways, such as changing the direction or speed of an object, causing objects to move closer or further away from each other, and creating forces between objects.

5. What are some real-life examples of relative acceleration?

Some examples of relative acceleration in everyday life include a car accelerating past another car on a highway, a rocket accelerating towards the moon, and a person riding a bike and pedaling faster to catch up to another person walking ahead of them.

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