Another Physics Problem?

Question

There are two cars: One at rest, the other at a constant velocity of 29 m/s. Car A, which has V=29m/s passes Car B. Car B begins to accelerate at a uniform rate of 2.8m/s/s as soon as Car A passes. What distance does Car B travel before passing Car A?

Answer 1

Let the problem start as soon as Car A passes Car B.  The equation of motion for Car B is

x = vt = 29t

The equation of motion (position) for Car A is

x = v0t + (1/2) a t^2 = (1/2)(2.8)t^2 = 1.4t^2

When Car A passes Car B, obviously they'll be at the same place, so the x's will be the same.  Equate and solve:

29t = 1.4t^2

Note that one of your solutions will be t = 0.  We already knew that - we chose to start the problem when Car B passed Car A, and obviously they will be at the same place then.  The other solution is the one you want.