Increase in speed, decrease in time...

keinekatze

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 4.0 mi/h, the time to go one mile decreases by 9 s. What was your original speed in mi/h?

(this is from chapter 1 in my physics book, but it's been a long time since I've had math and for the life of me can't figure out how to set it up!)

Xmph=original speed
t= original time seconds/mile
s= seconds
It seems to me that Xmph/t = (xmph+4mph)/(t-9s), but I can't figure out where to move from here... please, any help on the set-up of this equation would be greatly appreciated! I know this is supposed to be easy but my brain just can't work it...

Gamma

From the first hint, you are trevalling one mile in t seconds.

So your original speed is v = 1/t miles/s

Second hint: New speed = v + 4 = (1/t + 4) miles/s ---------(1)

Time to go 1 mile decreesed by 9 sec means it took (t-9) sec to travel one mile this time.

So new speed can also be written as 1/(t-9) ------(2)

combine (1) and (2).

1/t + 4 = 1 /(t-9)

Gamma

Solve for t. Your original speed is 1/t miles/s

caseys

Please forgive me for trying to exercise my 50+ year old mind to keep alzheimer's at bay...by working on physics problems.

I came across this problem and tried to solve for "t" ending up with t= 37 seconds. It was at this point that I got a brain "phart". (sorry)

I am lost. What am I not thinking about. Just a hint to get me back on track...I want to think this through myself if I can. Thank you. Casey

caseys

...after sleeping on this I got it figured out.