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
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.
1 answer