Please Check Answer -- Arrival times of 2 cars....

  • Thread starter Julie
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You got the idea, but messed up with the units. You need to keep the units consistent. You don't need to convert the distance to meters. Just be consistent with your units.
  • #1
Julie

Homework Statement


Two cars travel in the same direction along a straight highway. One at a constant speed of 23m/s and the other at 29m/s. How much sooner does the faster car arrive at the destination 16 km away?

Homework Equations


Please check answer

The Attempt at a Solution


T23 = 9.942 miles/23 mph = .432 hours .432 hour=60 minutes/hour .432(60) = 25.92 minutes

T29 = 9.942 miles/29mph = .343 hours .343 hour = 60 minute/hour .343(60) = 20.57 minutes

25.92 minutes – 20.57 minutes = 5.35 minutes sooner
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  • #2
Since when is m/s the same as mph?
 
  • #3
Thanks... My (probably wrong) logic says I would need to convert the kilometers into meters, making 16km = .016 meters. But this gives me a weird answer

t23=.016/23 = .0006
t29 = .016/29 = .0005

I seriously doubt this is correct. Can you give me a clue which part of the problem I need to convert?
 
  • #4
Julie said:
Thanks... My (probably wrong) logic says I would need to convert the kilometers into meters, making 16km = .016 meters. But this gives me a weird answer

t23=.016/23 = .0006
t29 = .016/29 = .0005

I seriously doubt this is correct. Can you give me a clue which part of the problem I need to convert?
Do you really think the 16 km is the same distance as 1.6 cm?
 
  • #5
I tried converting m/s into kilometers/hour using 1 m/s = 18/5 km/hr so 23 (18) = 414/5=82.8 kilometers/hour
29(18) = 522/5 = 104.4 kilometers/hour

16/82.8 = .19 * 60 = 11.4 minutes
16/104.4 = .15 * 60 = 9.2 minutes

11.4 - 9.2 = 2.2 minutes faster
 
  • #6
Julie said:
I tried converting m/s into kilometers/hour using 1 m/s = 18/5 km/hr so 23 (18) = 414/5=82.8 kilometers/hour
29(18) = 522/5 = 104.4 kilometers/hour

16/82.8 = .19 * 60 = 11.4 minutes
16/104.4 = .15 * 60 = 9.2 minutes

11.4 - 9.2 = 2.2 minutes faster
Good. Now try to solve it in terms of meters, rather than km. Hint: 16 km = 16000 meters.
 
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  • #7
The key here - and it seems you were on the right track - is to ask yourself if the numbers make sense intuitively.
Always check that the intermediate steps and the answer make sense.

Would a car traveling at about 100km/h take about 11 minutes to travel 16km? If it did that for an hour, would that be about ~6 times farther in ~6 times as long?

If one is moving at about 100, and the other is moving at about 80, would you expect that the faster one would take about 20% less time? Is ~9 minutes close to 20% shorter than 11 minutes?

Does the answer make sense intuitively?

Maths is not merely about the formulae and the numbers, it is about 'getting' it ('groking' it if you're old enough).
 
  • #8
ok... 16 km = 16000 meters
23 m/s = 82,800 meters per hour
29 m/s = 104,400 meters per hour

16000/82800 = .19 hour *60 = 11.4
16000/104400 = .15 hour *60 = 9.2
11.4 - 9.2 = 2.2 minutes faster
 
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  • #9
Awesome.

Another great tactic (if you're not being timed) is to try the problem a different way and see if you get the same answer. :wink:
 
  • #10
Julie said:
ok... 16 km = 16000 meters
23 m/s = 82,800 meters per hour
29 m/s = 104,400 meters per hour

16000/82800 = .19 hour *60 = 11.4
16000/104400 = .15 hour *60 = 9.2
11.4 - 9.2 = 2.2 minutes faster
How about not using hours at all? Get the answer in seconds, and then divide by 60 to get minutes.
 
  • #11
Dave and Chestermiller,

Thanks! Trying the problem a couple of different ways is a great approach. Is my answer correct? This is a practice problem I need to understand for an upcoming quiz.
 
  • #12
Chestermiller - I am getting a different answer when solving this way. If I am not doing anything wrong, which is the correct answer?

So it would be 16000/23 = 695.65/60 = 11.59 minutes
16000/29 = 551.72/60 = 9.20 minutes
11.59 - 9.20 = 2.39 minutes
 
  • #13
Yes, in your other answers, your intermediate steps did some rounding.
For example, 16,000/82,600 actually equals a decimal with more significant digits than my calculator allows. You rounded to 0.19.
Same with .15.

This is why Chester is pointing you at a better method, where you don't have to approximate any numbers in the intermediate steps.

Ideally, your work, and thus your answer, should use the same units as those you were given (metres, seconds), so that you don't introduce these kinds of errors.

You can always add an extra line to convert the final answer to a more readable unit (i.e. seconds to minutes).
 
Last edited:
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  • #14
Julie said:
Thanks... My (probably wrong) logic says I would need to convert the kilometers into meters, making 16km = .016 meters. But this gives me a weird answer

t23=.016/23 = .0006
t29 = .016/29 = .0005

I seriously doubt this is correct. Can you give me a clue which part of the problem I need to convert?

You need to distinguish between "kilo" and "milli". Your 0.016 m equals 16 mm, and also equals 16/1,000,000 km---quite a difference!
 
  • #15
Ray Vickson said:
You need to distinguish between "kilo" and "milli". Your 0.016 m equals 16 mm, and also equals 16/1,000,000 km---quite a difference!
Chester's post #4.
 

1. What is the purpose of checking the arrival times of 2 cars?

The purpose of checking the arrival times of 2 cars is to determine the time difference between their arrivals, which can provide valuable information about their speed, distance, and relative position.

2. How do you measure the arrival times of 2 cars?

The arrival times of 2 cars can be measured using a stopwatch or a timing device, such as a radar gun or a speedometer. It is important to start the timer at the same time for both cars in order to get accurate results.

3. What factors can affect the accuracy of arrival time measurements?

The accuracy of arrival time measurements can be affected by various factors, such as human error in starting or stopping the timer, weather conditions, traffic, and the speed and distance of the cars being observed.

4. How can the arrival times of 2 cars be used in scientific research?

The arrival times of 2 cars can be used in scientific research to study the physics of motion, including concepts like velocity, acceleration, and displacement. This information can also be used to analyze the performance of vehicles and improve transportation systems.

5. What are some potential applications of studying the arrival times of 2 cars?

The study of arrival times of 2 cars has various potential applications, such as in traffic engineering to improve traffic flow and reduce congestion, in sports to analyze the performance of athletes, and in accident reconstruction to determine the cause and sequence of events in a car crash.

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