Post a New Question


posted by .

Car A is moving at a constant speed of 10.0 m/s and passes Car B which is at rest. 10seconds after, the latter follows with an acceleration of 0.5 m/s^2. Determine the distance that the latter has to travel before it overtakes the former.

  • Physics -

    at time t seconds after passing car B,

    car A has gone 10t
    car B has gone 1/2 a(t-10)^2 = .25(t-10)^2

    the two distances are equal at time t=58.28

    so, both have gone 582.8m

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question