Simple physics kinematic one dimension problem?

  • Thread starter papi
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  • #1
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Homework Statement




Driving along a crowded freeway, you notice that it takes a time t to go from one mile marker to the next. When you increase your speed by 5.0 mi/h, the time to go one mile decreases by 13 s. What was your original speed in mi/hr?

Homework Equations



X (which in this case is 1 bc 1 mile)=VT

The Attempt at a Solution


Ive tried countless times on paper
I know you convert the 13s into hr by dividing 13/3600
i know you do something with t-13
but im so lost.
please help!
 

Answers and Replies

  • #2
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For consistency with metrics, it would be best to convert your units to metric format. Assuming there is no acceleration, the original velocity would be vi. Your displacement, delta(x), would then be equal to vi*t, t being time. When your velocity is increased, it takes 13s less to cover to cover one mile. Your velocity is increased by 5.0 mi/h, your time is 13s less, but your displacement is the same. With this, you should be able to conceptualize what to do (remember to convert your units).
 
  • #3
Redbelly98
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X (which in this case is 1 bc 1 mile)=VT
That's a good start. And you can rewrite that equation with the following changes:

If the velocity was V at first, what is the velocity after it increases by 5.0 mi/hr?

And if the time was T at first, what is the time after it decreases?

And yes, you'll need to be consistent with the time units, so the 13s becomes (13/3600) hr.

I recommend keeping all the length units as miles, since the given information uses miles and it says to give the answer in miles/hr.
 

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