Found 3 solutions by josgarithmetic, ikleyn, MathTherapy:
Answer by josgarithmetic(39799) (Show Source): You can put this solution on YOUR website!
Typical traveling rates catchup problem.
SPEEDS TIMES DISTANCES
SLOW EARLY 40 t+3 d
FAST LATE 55 t d
PURELY IN VARIABLES
SPEEDS TIMES DISTANCES
SLOW EARLY r t+h d
FAST LATE R t d
Use the given values to evaluate time t.
Answer by ikleyn(53763) (Show Source): You can put this solution on YOUR website!
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The first car is going 40 mi/h,and the second car is going 55 mi/h.the first car left 3 hours before the second car.
what equation could you use to find how many hours it will take for the second car to catch up to the first car?
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Everything is in these 4 lines below.
Time to catch = = = 8 hours.
The numerator 3*40 = 120 kilometers is the distance for the first car to be ahead at the moment the second car started.
The difference (55-40) in the denominator is the rate of decreasing the distance between two cars.
It is their relative speed.
That's all.
You should be able to write such an equation instantly as you read and understand the problem.
Producing tables is not the way solving such problems.
Answer by MathTherapy(10809) (Show Source): You can put this solution on YOUR website!
The first car is going 40mi/h,and the second car is going 55mi/h.the first car left 3 hours before the second car.what equation could you use to find how many hours it will take for the second car to catch up to the first car?
With T being time for the faster to catch up to the slower car, we get the following DISTANCE equation: 40(T + 3) = 55T
Solve this to get:
That's all.....NOTHING COMPLEX!!
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