SOLUTION: Mariel and Liu are running a long-distance race. Mariel runs at a rate of 5.4 m/s, and Liu runs at a rate of 7.6 m/s. If Mariel has a 20 m head start, how long will it take for Liu

Question 1185672: Mariel and Liu are running a long-distance race. Mariel runs at a rate of 5.4 m/s, and Liu runs at a rate of 7.6 m/s. If Mariel has a 20 m head start, how long will it take for Liu to catch Mariel?
Found 3 solutions by josgarithmetic, ikleyn, Theo:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
The speed of approach Liu to Mariel is 7.6-5.4=2.2 m/s.

x, time in seconds to close the 20 m difference:

Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.
Mariel and Liu are running a long-distance race. Mariel runs at a rate of 5.4 m/s,
and Liu runs at a rate of 7.6 m/s. If Mariel has a 20 m head start, how long will it take for Liu
to catch Mariel?
~~~~~~~~~~~~~~~

Let t be the time from their starting moment. Mariel run the distance 5.4 * t meters. Liu run the distance 7.6 * t meters. The difference of the distances is 20 m, so you write this equation 7.6*t - 5.4t = 20 meters. From the equation (7.6 - 5.4)t = 20 2.2 * t = 20 t = 20/2.2 = 9.0909... seconds = 9.1 seconds (rounded). ANSWER.

Solved.

In other form solution, the difference in their speeds is 7.6 - 5.4 = 2.2 m/s. It is the rate, at which the distance between them is decreased (the approaching speed). So the catching time is = 9.1 m/s. It is the same logic works as in Algebra solution above, so now you know two ways solving problem.

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    - Travel and Distance problems for two bodies moving in the same direction (catching up)
in this site.

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Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
mariel funs at a rate of 5.4 meters per second.
liu runs at a rate of 7.6 meters per second.
mariel has a 20 meter head start.

rate * time = distance.

when liu catches up to mariel, they will have both ran the same time;.
mariel runs a distance of x.
liu runs a distance of x + 2-

the equation for mariel is rate * time = distance becomes 5.4 * T = x
the equation for liu is rate * time = distance becomes 7.6 * T = x + 20

solve both equations for T to get:

T = x/5.4 for mariel.
T = (x + 20)/7.6 for liu.

since they're both equal to T, set them equal to each other to getg.

x/5.4 = (x + 20)/7.6

cross multiply to get:

7.6 * x = 5.4 * (x + 20)

simplify to get:

7.6 * x = 5.4 * x + 108

subtract 5.4 * x from both sides of the equation to get:

2.2 * x = 108

solve for x to get:

x = 108/2.2 = 49.090909091 meters.

x + 20 is therefore equal to 69.090909091 meters.

in 9.090909091 seconds, mariel traveled that * 5.4 meters per second = 49.090909091 meters.

in 9.090909091 seconds, liu traveled that * 7.6 meters per second = 69.090909091 meters.

since mariel had a 20 meter headstart, liu caught up to him in 9.090909091 seconds.

69.090909091 minus 49.090909091 = 20 meters.

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