Calculate Average Acceleration for the Detonator Ride at Worlds of Fun

In summary: When you multiply anything by 1 you get the same thing back.In summary, the problem involves finding the average acceleration of passengers on a ride at Worlds of Fun in Kansas City. The passengers undergo an acceleration from zero to 45 mi/h in 1.0 seconds. The answer is found by converting units and using the definition of acceleration, giving a final answer of 20.1 m/s^2. This answer may not have been obvious without knowledge of the unit conversion, but it is a reasonable unit for expressing acceleration in this context.
  • #1
nissanfreak
11
0
Equations of Motion Help!

I have been working on this for awile and still can't figure it out. What equation would I use to solve this problem? Here is the problem:

On a ride called the Detonator at Worlds of Fun in Kansas City, passengers accelerate straight downward from zero to 45 mi/h in 1.0 seconds. What is the average acceleration of the passengers on this ride?

Any and all help would be greatly appreciated! :smile:
 
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  • #2
This is a direct application of the definition of acceleration. No other equations are needed. You might need to do a unit conversion to match your answer to a given answer. The natural units that come from the information given would be miles per hour per second, but a book answer might be feet per second per second or feet per second squared. Other possibilities exist.
 
  • #3
The answer in the back of the book is 20.1 m/s^2 I just don't know how they got that? Can you help me figure this out! Thanks in advance!
 
  • #4
nissanfreak said:
The answer in the back of the book is 20.1 m/s^2 I just don't know how they got that? Can you help me figure this out! Thanks in advance!
It's just unit conversions. The acceleration is 45 miles/hour/second, and when converted, it gives you 20.1 meters/second^2
 
  • #5
First thing you're going to do is convert to metric units.

edit: I give up, it won't show up right.

Fill in hte proper conversion for each then multiply out. You'll get 20.1m/s^2
 
Last edited:
  • #6
Yea I was trying it out but it doesn't want to come out right. I believe I need to use one of the constant-acceleration equations of motion. But I am not sure which one? And how to plug the info into the equation?
 
  • #7
nissanfreak said:
Yea I was trying it out but it doesn't want to come out right. I believe I need to use one of the constant-acceleration equations of motion. But I am not sure which one? And how to plug the info into the equation?

I did it and got the correct answer. Well, my calculator did it. You are either using incorrect conversions or doing a conversion wrong.

1 mile = 1609.344 meters
1 hour = 3600 seconds
 
  • #8
can you show me how you multiplied it? I really appreciate it!
 
  • #9
45 mile/(hr*s) * 1 hr/3600 s = 0.0125 mile/s^2

.0125 mi/s^2 * 1609.344 m/1 mile = 20.1 m/s^2
 
  • #10
Thank you for your help! I do however have one last question. Would you be able find this answer 20.1m/s^2 if you didnt know that it was the answer? Sorry if this sounds dumb but I have only had 4 physics classes. I am taking a 5 week course this summer over physics so I am fairly new to all of this. Thanks again everyone for the help! :)
 
  • #11
nissanfreak said:
Thank you for your help! I do however have one last question. Would you be able find this answer 20.1m/s^2 if you didnt know that it was the answer? Sorry if this sounds dumb but I have only had 4 physics classes. I am taking a 5 week course this summer over physics so I am fairly new to all of this. Thanks again everyone for the help! :)

Yes, I definitely could have. All it is is a goofy unit conversion problem. One thing I learned from my AP physics class is that you need to keep with it and ask for help if you need it. Once you get the concepts, physics is really cool and a lot of fun. Until then...let's just say i still have bruises from banging my head against my desk. :tongue:
 
  • #12
nissanfreak said:
Thank you for your help! I do however have one last question. Would you be able find this answer 20.1m/s^2 if you didnt know that it was the answer? Sorry if this sounds dumb but I have only had 4 physics classes. I am taking a 5 week course this summer over physics so I am fairly new to all of this. Thanks again everyone for the help! :)

Yes you would be able to find that answer, but there is no way you would have known that you were supposed to express it in that form unless the question told you to do so. Miles per hour per second is a prefectly reasonable unit for expressing acceleration. It is the natural unit when speed is given in miles per hour and changing rapidly, such as for objects like the one in your problem or accelerating cars like dragsters. Don't feel bad about not anticipating the form of the answer you were given if the problem did not tell you to express it that way. Just make sure you understand how to do unit conversions when required.

You should notice that WhirlwindMonk accomplished the unit conversion by a series of multiplications by factors equal to 1. All unit conversions should be done that way. The ratio of two equal quantities is always 1. All you need to do is write the fraction in the form that is going to cancel the units you want replaced and leave you with the units you want. Sometimes it takes several steps to accomplish that. I am going to rewrite what you were already told, but in more detail, to emphasize this point.

[tex]
a = \frac{{45\frac{{mi}}{h}}}{{1s}} = \frac{{45mi}}{{hs}}\left( {\frac{{5280ft}}{{mi}}} \right)\left( {\frac{{12in}}{{ft}}} \right)\left( {\frac{{2.54cm}}{{in}}} \right)\left( {\frac{{1m}}{{100cm}}} \right)\left( {\frac{1 h}{{60\min }}} \right)\left( {\frac{{1\min }}{{60{\mathop{\rm s}\nolimits} }}} \right) = 20.1\frac{m}{{s^2 }}
[/tex]

Every fraction in parentheses has the value 1 because the numerator and denominator are equal.
 

1. What are the equations of motion?

The equations of motion are a set of mathematical equations that describe the motion of an object in terms of its position, velocity, acceleration, and time.

2. What is the difference between the three equations of motion?

The three equations of motion are the kinematic equations that relate an object's position, velocity, acceleration, and time. The first equation is used for motion with constant acceleration, the second equation is used for motion with constant acceleration and an initial velocity, and the third equation is used for motion with constant acceleration and an initial position.

3. How are the equations of motion used in real life?

The equations of motion are used in various fields of science and engineering, such as physics, mechanics, and aerospace, to analyze and predict the motion of objects. They are also used in everyday situations, such as calculating the time it takes to travel from one place to another or determining the speed of a moving vehicle.

4. What are the units used in the equations of motion?

The units used in the equations of motion depend on the system of measurement being used. In the SI (International System of Units), the units for position, velocity, and acceleration are meters, meters per second, and meters per second squared, respectively. Time is measured in seconds.

5. Can the equations of motion be applied to non-uniform motion?

Yes, the equations of motion can be applied to non-uniform or non-constant motion, but additional equations or modifications may be needed to accurately describe the motion. In such cases, calculus may be used to find the equations of motion.

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