A who travels 3 miles an hour, starts from a certain place five hours in advance of B who travels 4 miles an hour in the same direction. How many hours must B travel to overtake A?
When the distance traveled by A and B are equal then they have met and B then overtakes. Use d=rt (distance = rate times time.
3t = 4(t-5) solve
Note that 3t is distance of A and 4(t-5) is the distance of B.
So how do I solve it?
Are you an algebra student?
yes, why? I know that distance = rate/time
Use algebra and isolate t. Use algebraic rules and get t on the left and its' value on the right: t = ?? Hint clear the parenthesis of the equation provided by @sourwing
3t = 4t-5
and it is distance equals rate times time not divided by time.
Now subtract 4t from both sides of the equal sign. also correct your error when you cleared the parenthesis.
You should of gotten 3t = 4t - 20
3t - 4t = 4t -4t -20
-t = -20 Now divide both sides by -1
Where did you get 20 from?
Never mind
It will take 20 hours to overtake A.
Final results t = 20 hours. Should verify. Verification A distance 3 times 20 or 60 miles. B distnace 4(20-5) = 60 miles Yup, both traveled 60 miles.........good answer.
Yes, cassieblake, 20 hours is good. And good luck with your studies.
Thanks, Do you think you can help me with another problem?
I can try.
Say the rise is 4 and the run is 2. If I were to increase both the rise and run by 1, would it be the same rate of change? Why or why not?
No, the rate of change would not remain the same. Here is why the original rise over run (rise/run) is 4/2 or 2. Now if we add 1 to both the rise (getting5) and the run (getting 3) or rise over run is 5/3. 5/3 does not = 2 the resulting change is at a lower rate.
You follow ??
okay, thanks.
You're welcome.
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