# Two runners

Two runners ran simultaneously towards each other from locations distant 34.6 km. The average speed of the first runner was 1/5 higher than the average speed of the second runner.

How long should each ran a 34.6 km, if you know that they meet after 67 minutes?

Result

t1 =  122.8 min
t2 =  147.4 min

#### Solution:

$t = 67 \ min \ \\ s_1 + s_2 = 34.6 \ km \ \\ v_1 = v_2 (1+1/5) = 1.2 v_2 \ \\ \ \\ v_1 t + v_2 t = 34.6 \ \\ v_1+v_2 = 34.6/67 = 0.5164 \ km/min \ \\ 2.2 v_2 = 0.5164 \ \\ v_2 = 0.2347 \ km/min \ \\ v_1 = 1.2 v_2 = 0.2817 \ km/min \ \\ \ \\ t_1 = s/v_1 = 34.6/0.2817 = 122.8 \ \text{min}$
$t_2 = s/v_2 = 34.6/0.2347 = 147.4 \ \text{min}$

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