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two bikers left at the same time traveling opposite directions. The first biker was traveling 15 mph. The second biker was traveling 20mph. How long before they are 140 miles apart?

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The first biker travels distance d1 at speed s1 in t1 time. The second biker travels distance d2 at speed s2 in t2 time. In general, s = d/t, so d = st d1 = s1*t1, and d2 = s2*t2 d1 + d2 = s1*t1 + s2*t2 For them to be 140 miles apart means that the sum of their distances, d1 + d2 is 140 miles, or d1 + d2 = 140. Also, they travel the same amount of time, so you want t1 = t2 = t. You want d1 + d2 = 140, and t1 = t2 = t, so you have 140 = 15t + 20t 35t = 140 t = 4 4 hours

|dw:1363059098511:dw| For the first biker, we have: \[x_1=v_1t+x_0\] And second biker: \[x_2=v_2t+x_{0}\] From the drawing we also have: \[x_1-x_2=140\] Now, in the above equation, replace x_1 and x_2 with their respective formula: \[(v_1t+x_0)-(v_2t+x_{0})=140\] Plug in the value's of v_1(15), v_2(20) and x_0(0) and you have t: the time in which the two bikers are 140 miles apart!

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