Question

Two particles A and B are moving in the same direction on parallel horizontal tracks. At a a certain point the particle A travelling with a speed of 7ms^-1 and accelerating uniformly at 1.5ms^-2 overtakes B travelling at 3ms^-1 and accelerating uniformly at 2.5ms^-2. Calculate the period of time which elapses before B overtakes A. If, after this time, B then ceases to accelerate and continues at constant speed, calculate the rime taken for A to overtake B again.

Answer 1

You must set up this equation for both particles: \[position(final)=position(initial)+velocity(initial)+1/2*acell*time^2\]You can assume that position(initial) is zero for both particles and that the position(final) is the same for both particles when they're crossing. because their final positions are the same, just set both equations equal to each other and solve for time. It will look like\[time=\sqrt{\frac{(velocityA(initial)-velocityB(initial))}{(1/2accelB-1/2accelA)}}\]
Well that's how I think it works! For the second part of the problem you have to solve for the velocity of particle B at the time you solved for in part a using one (or more) of these kinematic equations shown in the link below:
http://www.physicsclassroom.com/Class/1DKin/U1L6a1.gif

Kind of long explanation, but hoped I helped!

Answer 2

i plugged in the givens and still the answers is not the same as found in the back of my book.