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10 March, 19:53

Drivers a and b travel west from boston. driver a leaves three hours earlier traveling at a constant speed of 68mph. driver b follows at a constant speed of 85 mph. how many hours would it take for driver b to catch up to driver a?

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  1. 10 March, 20:24
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    We can answer this using one of the equations of linear motion:

    v = d / t

    where:

    v = velocity

    d = distance

    t = time

    In the problem, we are asked to find for the time in which Driver B will catch up to Driver A. Therefore, find the time when dA = dB. Rearranging the equation and equation dA and dB will result in:

    vA * tA = vB * tB - - - > 1

    It was given that:

    vA = 68 mph

    tA = tB + 3 (since person A was travelling 3 hours earlier)

    vB = 85 mph

    tB = unknown

    Substituting into equation 1:

    68 * (tB + 3) = 85 * tB

    68 tB + 204 = 85 tB

    tB = 12 hrs

    Therefore driver B would catch up to driver A after 12 hrs.
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